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Wize University Calculus 1 Textbook > Integrals

Integration by Substitution (U-Substitution)

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Integration by Substitution (U-Substitution)

There is no "Chain Rule for Integrals" like there is for derivatives. When we are trying to integrate compositions of functions, often we use the technique of U Substitution.

U Substitution

If f(x)f\left(x\right) is a differentiable function with a continuous derivative f′(x)f'\left(x\right), by letting u=f(x)u=f\left(x\right)we have
∫abg(f(x))×f′(x)dx=∫f(a)f(b)g(u)du\boxed{\displaystyle \int_a^b g(f(x))\times f'(x)dx=\int_{f(a)}^{f(b)} g(u)du }

Wize Tip
We typically use U-Substitution when the derivative of one part shows up elsewhere in the integral.

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Procedure for U-Substitution

  1. Let uu equal part of the integrand (usually what's inside parenthesis or the "messy" part of the function)
  2. Differentiate uu to get dudu
  3. Solve for dxdx
  4. Substitute uu and dudu into your integral (for definite integrals replace the bounds)
  5. Integrate with respect to uu
  6. Substitute the original xxexpression back in (only for indefinite integrals)

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Example: U-Sub

Evaluate the following definite integral

∫−11(2−4x)(x−x2)5 dx\displaystyle\int_{-1}^1\left(2-4x\right)\left(x-x^2\right)^5\ dx.

1. Pick "u" expression

We have 2 candidates for uu: 2−4x2-4x and x−x2x-x^2
Both of these are wrapped inside brackets, let's pick the higher degree part: u=x−x2u=x-x^2


2. Differentiate u to get du=...dx
du=(1−2x) dxdu=\left(1-2x\right)\ dx


3. Solve for dx
dx=11−2x du\displaystyle dx=\frac{1}{1-2x}\ du


4. Substitute u and replace dx

(Since this is a definite integral, we need to replace the bounds)


When the lower bound is x=−1→u=−2x=-1\to u=-2
When the upper bound is x=1→u=0x=1\to u=0
Let's substitute in the uu expression, the bounds, and replace the dtdt expression:
=∫−20(2−4x)(u)5 (11−2xdu)\displaystyle =\int_{-2}^0\left(2-4x\right)\left(u\right)^5\ \left(\frac{1}{1-2x}du\right)

=∫−202(1−2x)(u)5 (11−2xdu)\displaystyle=\int_{-2}^02\left(1-2x\right)\left(u\right)^5\ \left(\frac{1}{1-2x}du\right)

=2∫−20(u)5du\displaystyle=2\int_{-2}^0\left(u\right)^5du


5. Integrate this new u expression
=2∫−20u5 du\displaystyle=2\int_{-2}^0u^5\ du

=2[u66]−20\displaystyle=2\left[\frac{u^6}{6}\right]_{_{-2}}^{^0}

=2[0−(−2)66]\displaystyle=2\left[0-\frac{\left(-2\right)^6}{6}\right]

=−643\displaystyle =-\frac{64}{3}

Wize Tip
Since this is a definite integral, we don't have to substitute the original x expression back in.

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Example: U Sub

Evaluate the following definite integral

∫14xx2+1 dx\displaystyle \int_{1}^{4}x\sqrt{x^2+1}\ dx

Notice that ddx(x2+1)=2x dx\frac{d}{dx}(x^2+1)=2x \ dx, which is similar to the x dxx\ dx that appears outside the square root.

1. Pick "u" expression
u=x2+1u=x^2+1

2. Differentiate u to get du
du=2x dxdu=2x\ dx

3. Solve for dx
 dx=12xdu\ dx=\frac{1}{2x}du

4. Substitute u and replace dx

(Since this is a definite integral, we need to replace the bounds)

u(1)=(1)2+1=2; u(4)=42+1=17u(1)=(1)^2+1=2; \ u(4)=4^2+1=17

So then

∫14xx2+1 dx=∫217xu⋅12x du\begin{array}{rl} \displaystyle\int_1^4 x\sqrt{x^2+1}\ dx & = \displaystyle\int_2^{17}x\sqrt{u} \cdot \frac{1}{2x} \ du \\[+2em] \end{array}
5. Integrate this new u expression

=12∫217u1/2 du=12[u3/23/2]217=12[2u3/23]217=12⋅23[u3/2]217=13(173/2−23/2)\begin{array}{rl} &=\dfrac{1}{2}\displaystyle\int_2^{17} u^{1/2}\ du \\[+2em] &= \dfrac{1}{2}\bigg[\dfrac{u^{3/2}}{3/2}\bigg]_2^{17}\\[+2em] &= \dfrac{1}{2}\bigg[\dfrac{2u^{3/2}}{3}\bigg]_2^{17}\\[+2em] &=\dfrac{1}{2}\cdot \dfrac{2}{3}\bigg[u^{3/2}\bigg]_2^{17}\\[+2em] &=\dfrac{1}{3}\left( 17^{3/2}-2^{3/2} \right) \end{array}

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Example: U-Sub (with back-substitution)

Evaluate the following indefinite integral

∫x−x31+x2 dx \displaystyle \int_{ }^{ }\frac{x-x^3}{1+x^2}\ dx

Simplifying will not make this integration any easier, let's try u-substitution.

Normally, we will pick u to be the numerator because the numerator is of degree 3 and denominator is of degree 2. However, if you try that, you'll realize that we have nothing to cancel out with in our 4th step. So instead, let's pick u to be the denominator.

1. Pick "u" expression
u=1+x2u=1+x^2

2. Differentiate u to get du
du=2x dxdu=2x\ dx

3. Solve for dx
dx=du2x\displaystyle dx=\frac{du}{2x}

4. Substitute u and replace dx
∫x−x31+x2 dx\displaystyle \int_{ }^{ }\frac{x-x^3}{1+x^2}\ dx

=∫x−x3u (du2x)\displaystyle=\int_{ }^{ }\frac{x-x^3}{u}\ \left(\frac{du}{2x}\right)

=∫x(1−x2)u (du2x)\displaystyle=\int_{ }^{ }\frac{x\left(1-x^2\right)}{u}\ \left(\frac{du}{2x}\right)

=∫(1−x2)u (du2)\displaystyle=\int_{ }^{ }\frac{\left(1-x^2\right)}{u}\ \left(\frac{du}{2}\right)

Since the expression still involves an x term, we need to back-substitute:
From u=1+x2u=1+x^2, we have x2=u−1x^2=u-1.
We can now use this to replace the x2x^2 in our expression:

=∫(1−(u−1))u (du2)\displaystyle=\int_{ }^{ }\frac{\left(1-\left(u-1\right)\right)}{u}\ \left(\frac{du}{2}\right)

=12∫2−uu du\displaystyle=\frac{1}{2}\int_{ }^{ }\frac{2-u}{u}\ du

=12∫[2u−1] du\displaystyle=\frac{1}{2}\int_{ }^{ }\left[\frac{2}{u}-1\right]\ du

5. Integrate this new u expression
=12[2ln⁡∣u∣−u]+C\displaystyle=\frac{1}{2}\left[2\ln\left|u\right|-u\right]+C

=ln⁡∣u∣−u2+C\displaystyle=\ln\left|u\right|-\frac{u}{2}+C

6. Sub the x expression back in for u:
=ln⁡∣1+x2∣−1+x22+C\displaystyle=\ln\left|1+x^2\right|-\frac{1+x^2}{2}+C

*Since 1+x2>01+x^2>0 for all values of x, we don't need the absolute value signs.

Practice: U-Sub

Evaluate the following indefinite integral

∫ln⁡(x)xdx\displaystyle\int\frac{\ln(x)}{x}d{x}


Practice: U-Sub

Evaluate the following indefinite integral

∫sec⁡3xtan⁡3x(1+sec⁡3x)3/2dx\displaystyle\int\frac{\sec 3x\tan 3x}{(1+\sec 3x)^{3/2}}d{x}

Practice: U-Sub

Evaluate the following indefinite integral

∫tan⁡xln⁡(cos⁡x)  ⁣dx\displaystyle\int\frac{\tan x}{\ln(\cos x)}\de{x}


Practice: U-Sub

Evaluate the following indefinite integral

∫1e3x−1dx\displaystyle\int\frac{1}{\sqrt {e^{3x}-1}}d{x}



Extra Practice
Integration by Substitution: Indefinite with Trig
Find the following integral: ∫1x2sin⁡(1x)cos⁡(1x) dx\displaystyle\int\frac{1}{x^2}\sin\left(\frac{1}{x}\right)\cos\left(\frac{1}{x}\right) \ dx.
Integration by Substitution: Indefinite with Power
Compute the following integral: ∫(x−6)(x+6)1/3 dx\displaystyle \int (x-6)(x+6)^{1/3}\,\text{d}x
Integration by Substitution
If ∫25f(2x) dx=6\int_2^5f\left(2x\right)\ dx=6 and ∫210f(x)=20\int_2^{10}f\left(x\right)=20, find ∫24f(x) dx\int_2^4f\left(x\right)\ dx.
Integration by Substitution
∫e1xx2dx\displaystyle \int_{ }^{ }\frac{e^{\frac{1}{x}}}{x^2}dx
Integration by Substitution
Evaluate ∫eex+xdx\int_{ }^{ }e^{e^x+x}dx.
Integration by Substitution
Evaluate the definite integral ∫0π/3ln⁡(cos⁡x)tan⁡x dx{\displaystyle \int_{0}^{\pi/3}}\ln{(\cos x)}\tan x \space dx.
Practice Question: U-Sub *
Compute∫1e3x−1  ⁣dx{\displaystyle\int}\frac{1}{\sqrt {e^{3x}-1}}\de{x}.
Compute∫1e3x−1  ⁣dx{\displaystyle\int}\frac{1}{\sqrt {e^{3x}-1}}\de{x}.
Practice Question
Compute∫1e3x−1  ⁣dx{\displaystyle\int}\frac{1}{\sqrt {e^{3x}-1}}\de{x}.
Practice Question
Compute∫1e3x−1  ⁣dx{\displaystyle\int}\frac{1}{\sqrt {e^{3x}-1}}\de{x}.
Practice Question
Compute∫1e3x−1  ⁣dx{\displaystyle\int}\frac{1}{\sqrt {e^{3x}-1}}\de{x}.
Practice Question
Compute∫1e3x−1  ⁣dx{\displaystyle\int}\frac{1}{\sqrt {e^{3x}-1}}\de{x}.
Practice Question: U-Sub *
Practice Question
Compute∫1e3x−1  ⁣dx{\displaystyle\int}\frac{1}{\sqrt {e^{3x}-1}}\de{x}.
Practice Question: U-Sub *
Practice Question
Compute∫1e3x−1  ⁣dx{\displaystyle\int}\frac{1}{\sqrt {e^{3x}-1}}\de{x}.
Integration by Substitution
Evaluate the definite integral ∫0π/3ln⁡(cos⁡x)tan⁡x dx{\displaystyle \int_{0}^{\pi/3}}\ln{(\cos x)}\tan x \space dx.
Integration by Substitution: Definite with Exponential
Compute the following integral: ∫13xe2x25 dx\displaystyle \int_{1}^{3}\frac{xe^{2x^2}}{5}\,\text{d}x
Integration by Substitution
Evaluate the definite integral ∫0π/3ln⁡(cos⁡x)tan⁡x dx{\displaystyle \int_{0}^{\pi/3}}\ln{(\cos x)}\tan x \space dx.
Substitution with Definite Integral
Practice: Substitution with Definite Integral
Evaluate ∫1e(ln⁡x−5)2xdx\displaystyle \int_1^e\frac{\left(\ln x-5\right)^2}{x}dx.
Substitution with Definite Integral
Practice: Substitution with Definite Integral
Evaluate ∫1e(ln⁡x−5)2xdx\displaystyle \int_1^e\frac{\left(\ln x-5\right)^2}{x}dx.
Integration by Substitution
∫e1xx2dx\displaystyle \int_{ }^{ }\frac{e^{\frac{1}{x}}}{x^2}dx
Integration by Substitution
Evaluate the definite integral ∫0π/3ln⁡(cos⁡x)tan⁡x dx{\displaystyle \int_{0}^{\pi/3}}\ln{(\cos x)}\tan x \space dx.
Practice Question: U-Sub
Compute ∫ln⁡(x)x  ⁣dx.{\displaystyle\int}\frac{\ln(x)}{x}\de{x}.
Practice Question: U-Sub
Practice Question
Evaluate ∫sec⁡3xtan⁡3x(1+sec⁡3x)3/2  ⁣dx{\displaystyle\int}\frac{\sec 3x\tan 3x}{(1+\sec 3x)^{3/2}}\de{x}.
Integration by Substitution
Find the antiderivative of
f(x)=−cos⁡6x5sin⁡26xf(x)= -\frac{\cos 6x}{5\sin^2 6x}
Integration by Substitution
Evaluate ∫eex+xdx\int_{ }^{ }e^{e^x+x}dx.
Integration by Substitution
∫e1xx2dx\displaystyle \int_{ }^{ }\frac{e^{\frac{1}{x}}}{x^2}dx
Integration by Substitution: Definite with Exponential
Compute the following integral: ∫13xe2x25 dx\displaystyle \int_{1}^{3}\frac{xe^{2x^2}}{5}\,\text{d}x
Substitution with Definite Integral
Evaluate ∫1e(ln⁡x−5)2xdx\displaystyle \int_1^e\frac{\left(\ln x-5\right)^2}{x}dx.
Integration by Substitution
Evaluate the definite integral ∫0π/3ln⁡(cos⁡x)tan⁡x dx{\displaystyle \int_{0}^{\pi/3}}\ln{(\cos x)}\tan x \space dx.
Evaluate the integral ∫x3(1+x4)14dx\displaystyle \int_{ }^{ }x^3(1+x^4)^{\frac{1}{4}}dx.
Find the indefinite integral ∫x1+x2dx\int\frac{x}{1+x^2}dx.
Evaluate the integral I=∫cos⁡(x+1)x+1dx\displaystyle I=\int\frac{\cos(\sqrt{x+1})}{\sqrt{x+1}}dx.
Evaluate the integral ∫01x3⋅3x4+2 dx\displaystyle\int_0^1x^3\cdotp3^{x^4+2}\ dx.
Evaluate the integral ∫cos⁡x sec⁡2(sin⁡x)dx\displaystyle\int_{ }^{ }\cos x\ \sec^2\left(\sin x\right)dx.
Evaluate the integral ∫cos⁡x sec⁡2(sin⁡x)dx\displaystyle\int_{ }^{ }\cos x\ \sec^2\left(\sin x\right)dx.
Evaluate the integral ∫cos⁡x sec⁡2(sin⁡x)dx\displaystyle\int_{ }^{ }\cos x\ \sec^2\left(\sin x\right)dx.
Evaluate the integral ∫cos⁡x sec⁡2(sin⁡x)dx\displaystyle\int_{ }^{ }\cos x\ \sec^2\left(\sin x\right)dx.
Evaluate the integral ∫1arcsin⁡x1−x2dx\displaystyle\int_{ }^{ }\frac{1}{\arcsin x\sqrt{1-x^2}}dx. (If you need to input an inverse trigonometric function, the computer asks that you use the "arc" notation instead of "-1", i.e. arcsin⁡\arcsin instead of sin⁡−1\sin^{-1}.)
Find the definite integral ∫14exxdx\displaystyle\int_1^4\frac{e^{\sqrt{x}}}{\sqrt{x}}dx.
Practice Question: U-Sub
Compute ∫ln⁡(x)x  ⁣dx.{\displaystyle\int}\frac{\ln(x)}{x}\de{x}.
Complete the Square, Tan, Split the Fraction: Integration by Substitution
Evaluate ∫x(x2−x+1)2 dx\displaystyle\int\frac{x}{(x^2-x+1)^2} \ dx
Practice: Integration by Substitution (with back-substitution)
Practice: Integration by Substitution (with back-substitution)
Find ∫2x32x2+1dx\int_{ }^{ }\frac{2x^3}{2x^2+1}dx.
Integration by Substitution
∫e1xx2dx\displaystyle \int_{ }^{ }\frac{e^{\frac{1}{x}}}{x^2}dx
Practice: Integration by Substitution
Practice: Integration by Substitution
Evaluate ∫02(x−2)e(x2−4x)dx\int_0^2\left(x-2\right)e^{\left(x^2-4x\right)}dx.
Practice: Integration by Substitution (~F2018 Final Q36)
Practice: Integration by Substitution
Find ∫cot⁡x ln⁡(sin⁡x)dx\int_{ }^{ }\cot x\ \ln\left(\sin x\right)dx.
Practice: Integration by Substitution (~F2017 Final Q35)
Practice: Integration by Substitution
Evaluate ∫1e(ln⁡x−5)2xdx\int_1^e\frac{\left(\ln x-5\right)^2}{x}dx.
Practice: Property of Definite Integral
Practice: Property of Definite Integral
Suppose f(x)f(x) is continuous and let ∫07f(x)dx=7\displaystyle \int_0^7f\left(x\right)dx=7 and ∫710f(x)dx=10\displaystyle \int_7^{10}f\left(x\right)dx=10.
Find 5∫02f(5x)dx\displaystyle 5\int_0^2f\left(5x\right)dx.
Complete the Square, Tan, Split the Fraction: Integration by Substitution
Evaluate ∫x(x2−x+1)2 dx\displaystyle\int\frac{x}{(x^2-x+1)^2} \ dx
Integration by Substitution: Definite with Exponential
Evaluate the integral: ∫01x3 3x4+2 dx\displaystyle\int_0^1x^3 \ 3^{x^4+2}\ dx
Integration by Substitution: Powers and Root
Compute the following integral: ∫ x3x2+1 dx\displaystyle \int\,x^3\sqrt{x^2+1} \ \text{d}x
Practice: Indefinite with Root
Q:\textbf{Q:} Evaluate ∫2xx2−1dx\displaystyle\int_{ }^{ }\frac{2x}{\sqrt{x^2-1}}dx
Integration by Substitution: Indefinite with Root
Compute the following integral: ∫31−4x2 dx\displaystyle \int\frac{3}{\sqrt{1-4x^2}}\,\text{d}x
Integration by Substitution: Indefinite with Fraction
Compute the following integral: ∫11+2x2 dx\displaystyle \int \frac{1}{1+2x^2} \ \text{d}x
Practice: Indefinite with Inverse Trig
Q.\textbf{Q.} Compute the following integral: ∫arcsinx1−x2 dx\displaystyle \int\frac{\text{arcsin}{x}}{\sqrt{1-x^2}}\,\text{d}x
Integration by Substitution: Indefinite with Square Root
Compute the following integral: ∫3x+2 dx\displaystyle \int \sqrt{3x+2}\,\text{d}x
Evaluate ∫1e(ln⁡x−5)4x dx\displaystyle\int_1^e\frac{\left(\ln x-5\right)^4}{x}\ dx
Suppose f(x)f(x) is continuous and let ∫07f(x)dx=7\displaystyle\int_0^7f\left(x\right)dx=7 and ∫710f(x)dx=10 \displaystyle\int_7^{10}f\left(x\right)dx=10 , find 5∫02f(5x)dx\displaystyle5\int_0^2f\left(5x\right)dx.
Find ∫cot⁡x ln⁡(sin⁡x)dx\displaystyle\int_{ }^{ }\cot x\ \ln\left(\sin x\right)dx
Integration by Substitution: Indefinite with Trig
Find the following integral: ∫1x2sin⁡(1x)cos⁡(1x) dx\displaystyle\int\frac{1}{x^2}\sin\left(\frac{1}{x}\right)\cos\left(\frac{1}{x}\right) \ dx.
Integration by Substitution: Indefinite with Exponentials
Compute the following integral: ∫ez+ez dz\displaystyle \int e^{z+e^z}\,\text{d}z
Integration by Substitution: Indefinite with Power
Compute the following integral: ∫(x−6)(x+6)1/3 dx\displaystyle \int (x-6)(x+6)^{1/3}\,\text{d}x
Practice: Definite with Roots
Q.\textbf{Q.} Evaluate ∫14xx2+1 dx\displaystyle \int_{1}^{4}x\sqrt{x^2+1}\,\text{d}x
Integration by Substitution: Definite with Exponential
Compute the following integral: ∫13xe2x25 dx\displaystyle \int_{1}^{3}\frac{xe^{2x^2}}{5}\,\text{d}x
Practice: Indefinite with Inverse Trig
Compute the following integral: ∫arcsinx1−x2 dx\displaystyle \int\frac{\text{arcsin}{x}}{\sqrt{1-x^2}}\,\text{d}x
(a) =[arctan⁡(x)]22+C=\frac{[\arctan(x)]^2}{2}+C (b) =[arcsin⁡(x)]22+C=\frac{[\arcsin(x)]^2}{2}+C (c) =121−x2+C=\frac12\sqrt{1-x^2}+C (d) diverges
Practice: Integration by Substitution
Practice: Integration by Substitution derived from Final
Practice: Substitution with Definite Integral
Practice: Substitution with Definite Integral
Evaluate ∫1e(ln⁡x−5)2xdx\displaystyle \int_1^e\frac{\left(\ln x-5\right)^2}{x}dx.
Practice: Integration by Substitution (with back-substitution)
Practice: Integration by Substitution (with back-substitution)
Find ∫2x32x2+1dx\int_{ }^{ }\frac{2x^3}{2x^2+1}dx.
Practice: Substitution with Definite Integral
Practice: Substitution with Definite Integral
Evaluate ∫1e(ln⁡x−5)2xdx\displaystyle \int_1^e\frac{\left(\ln x-5\right)^2}{x}dx.
Example: Integration by Substitution
Evaluate ∫1x+x1/2dx\displaystyle \int_{ }^{ }\frac{1}{x+x^{1/2}}dx
Practice: Integration by Substitution (~2019 Test1 A10)
Practice: Integration by Substitution derived from Midterm
Evaluate ∫e1xx2dx\displaystyle \int_{ }^{ }\frac{e^{\frac{1}{x}}}{x^2}dx
Practice: Substitution with Definite Integral
Practice: Substitution with Definite Integral
Evaluate ∫1e(ln⁡x−5)2xdx\displaystyle \int_1^e\frac{\left(\ln x-5\right)^2}{x}dx.
Integration by Substitution
∫tan⁡x⋅ln⁡(cos⁡x)dx=\displaystyle\int_{ }^{ }\tan x\cdot\sqrt{\ln\left(\cos x\right)}dx=
Integration by Substitution
∫e1x3x2dx=\displaystyle \int_{ }^{ }\frac{e^{\frac{1}{x}}}{3x^2}dx=
Definite Integrals: Integration by Substitution
If ∫−27g(x)dx=13\int_{-2}^7g\left(x\right)dx=13 and ∫572g(x)=10\int_5^72g\left(x\right)=10, then ∫−152g(2x)dx=\int_{-1}^{\frac{5}{2}}g\left(2x\right)dx=
Find ∫12exxdx\int_{ }^{ }\frac{1}{2e^{\sqrt{x}}\sqrt{x}}dx
Evaluate ∫−102x23−3x3dx\int_{-1}^0\frac{2x^2}{3-3x^3}dx.
Find ∫3x(x2−6)dx\displaystyle\int_{ }^{ }\frac{3x}{\sqrt{\left(x^2-6\right)}}dx.
Integration by Substitution
Evaluate the definite integral ∫0π/3ln⁡(cos⁡x)tan⁡x dx{\displaystyle \int_{0}^{\pi/3}}\ln{(\cos x)}\tan x \space dx.
Integration by Substitution
Evaluate the definite integral ∫0π/3ln⁡(cos⁡x)tan⁡x dx{\displaystyle \int_{0}^{\pi/3}}\ln{(\cos x)}\tan x \space dx.
Practice: Integration by Substitution (with back-substitution)
Practice: Integration by Substitution (with back-substitution)
Find ∫2x32x2+1dx\int_{ }^{ }\frac{2x^3}{2x^2+1}dx.
Substitution with Definite Integral
Practice: Substitution with Definite Integral
Evaluate ∫1e(ln⁡x−5)2xdx\displaystyle \int_1^e\frac{\left(\ln x-5\right)^2}{x}dx.
Practice: Integration by Substitution (~2019 Test1 A10)
Practice: Integration by Substitution
Evaluate ∫e1xx2dx\displaystyle \int_{ }^{ }\frac{e^{\frac{1}{x}}}{x^2}dx
Evaluate ∫sin⁡(ln⁡x)xdx\int_{ }^{ }\frac{\sin\left(\ln x\right)}{x}dx
Integration by substitution
Evaluate ∫02xx2−1dx\displaystyle\int_0^2 \dfrac{x}{x^2-1}dx
Practice: Property of Definite Integral
Suppose f(x) is continuous and let ∫07f(x)dx=7\int_0^7f\left(x\right)dx=7 and ∫710f(x)dx=10\int_7^{10}f\left(x\right)dx=10. Find 5∫02f(5x)dx5\int_0^2f\left(5x\right)dx
Integration by Substitution
Evaluate ∫eex+xdx\int_{ }^{ }e^{e^x+x}dx.
Integration by Substitution
Evaluate ∫13x2+1x3+3x dx\displaystyle \int_{1}^{3} \frac{x^2+1}{\sqrt{x^3+3x}} \,dx
Integration by substitution
Evaluate ∫01x31+x2 dx\displaystyle \int_{0}^{1} \frac{x^3}{\sqrt{1+x^2}} \,dx
Integration by Substitution
Find the antiderivative of
f(x)=−cos⁡6x5sin⁡26xf(x)= -\frac{\cos 6x}{5\sin^2 6x}
Practice Question: U-Sub
Practice Question
Compute ∫ln⁡(x)x  ⁣dx.{\displaystyle\int}\frac{\ln(x)}{x}\de{x}.
Evaluate ∫1xln⁡xdx\int_{ }^{ }\frac{1}{x\ln\sqrt{x}}dx
Find ∫x41+x10dx\int_{ }^{ }\frac{x^4}{1+x^{10}}dx
Evaluate ∫sin⁡(ln⁡x)xdx\int_{ }^{ }\frac{\sin\left(\ln x\right)}{x}dx
Evaluate ∫arctan⁡(1x)dx\int_{ }^{ }\arctan\left(\frac{1}{x}\right)dx
Evaluate ∫1xex+exdx\int_{ }^{ }\frac{1}{\sqrt{x}}e^{\sqrt{x}+e^{\sqrt{x}}}dx
Evaluate ∫12x2+23x3+2x+10dx\int_1^2\frac{x^2+\frac{2}{3}}{x^3+2x+10}dx
Evaluate ∫1ex−2dx\int_{ }^{ }\frac{1}{e^x-2}dx
Evaluate ∫1xln⁡xdx\int_{ }^{ }\frac{1}{x\ln\sqrt{x}}dx
∫141x+x3dx\int_1^4\frac{1}{\sqrt{x}+\sqrt{x^3}}dx
Practice Question: U-Sub *
Practice Question
Compute∫1e3x−1  ⁣dx{\displaystyle\int}\frac{1}{\sqrt {e^{3x}-1}}\de{x}.
Practice Question: U-Sub
Practice Question
Evaluate ∫sec⁡3xtan⁡3x(1+sec⁡3x)3/2  ⁣dx{\displaystyle\int}\frac{\sec 3x\tan 3x}{(1+\sec 3x)^{3/2}}\de{x}.
Integration Medley
Practice Question
Evaluate ∫tan⁡xln⁡(cos⁡x)  ⁣dx{\displaystyle\int}\frac{\tan x}{\ln(\cos x)}\de{x}.
Practice Question: U-Sub
Evaluate
∫sec⁡3xtan⁡3x(1+sec⁡3x)3/2dx{\displaystyle\int}\frac{\sec 3x\tan 3x}{(1+\sec 3x)^{3/2}}dx
Practice: Property of Definite Integral
Suppose f(x) is continuous and let ∫07f(x)dx=7\int_0^7f\left(x\right)dx=7 and ∫710f(x)dx=10\int_7^{10}f\left(x\right)dx=10. Find 5∫02f(5x)dx5\int_0^2f\left(5x\right)dx
Evaluate ∫1xln⁡xdx\int_{ }^{ }\frac{1}{x\ln\sqrt{x}}dx
Evaluate ∫ex∗3exdx\displaystyle \int e^x*3^{e^x}dx
Practice Question: U-Sub
Evaluate
∫sec⁡3xtan⁡3x(1+sec⁡3x)3/2dx{\displaystyle\int}\frac{\sec 3x\tan 3x}{(1+\sec 3x)^{3/2}}dx
Evaluate ∫(cosx)2sin(x)dx\displaystyle \int (cosx)2^{sin(x)}dx
Evaluate∫ln2ln3ex(ex−1)3 dx\text{Evaluate} \displaystyle \int_{ln2}^{ln3} e^x(e^x-1)^3 \space dx
Find ∫1(x)(lnx)dx\displaystyle \int \frac{1}{(x)(lnx)}dx
Practice Question: U-Sub *
Compute∫1e3x−1  ⁣dx{\displaystyle\int}\frac{1}{\sqrt {e^{3x}-1}}\de{x}.
Practice Question: U-Sub
Evaluate ∫sec⁡3xtan⁡3x(1+sec⁡3x)3/2  ⁣dx{\displaystyle\int}\frac{\sec 3x\tan 3x}{(1+\sec 3x)^{3/2}}\de{x}.
Integration Medley
Evaluate ∫tan⁡xln⁡(cos⁡x)  ⁣dx{\displaystyle\int}\frac{\tan x}{\ln(\cos x)}\de{x}.
Practice Question: U-Sub
Compute ∫ln⁡(x)x  ⁣dx.{\displaystyle\int}\frac{\ln(x)}{x}\de{x}.
Find ∫1−3x3+6x−9x2dx\displaystyle\int\frac{1-3x}{\sqrt{3+6x-9x^2}}dx
Complete the Square, Tan, Split the Fraction: Integration by Substitution
Evaluate ∫x(x2−x+1)2 dx\displaystyle\int\frac{x}{(x^2-x+1)^2} \ dx
Evaluate the integral ∫cos⁡(x+1)x+1dx\displaystyle \int\frac{\cos(\sqrt{x+1})}{\sqrt{x+1}}dx. Use C for the constant of integration.
Example: Integration by Substitution
Evaluate ∫1x+x1/2dx\displaystyle \int_{ }^{ }\frac{1}{x+x^{1/2}}dx
Practice: Properties of Definite Integral
Evaluate ∫−π2π2xcos⁡2x sin⁡(ex2) dx\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x\cos^2x\ \sin \left(e^{x^2}\right)\ dx.
Practice: Property of Definite Integral (~F2014 Final Q25) (~F2017 Final Q24)
Suppose f(x) is continuous and let ∫07f(x)dx=7\int_0^7f\left(x\right)dx=7 and ∫710f(x)dx=10\int_7^{10}f\left(x\right)dx=10. Find 5∫02f(5x)dx5\int_0^2f\left(5x\right)dx
Practice: Integration by Substitution
Evaluate ∫02(x−2)e(x2−4x)dx\int_0^2\left(x-2\right)e^{\left(x^2-4x\right)}dx.
Integration by Substitution
Find ∫cot⁡x ln⁡(sin⁡x)dx\int_{ }^{ }\cot x\ \ln\left(\sin x\right)dx.
Evaluate the integral I=∫cos⁡(x+1)x+1dx\displaystyle I=\int\frac{\cos(\sqrt{x+1})}{\sqrt{x+1}}dx.
Practice Question: U-Sub
Evaluate
∫sec⁡3xtan⁡3x(1+sec⁡3x)3/2dx{\displaystyle\int}\frac{\sec 3x\tan 3x}{(1+\sec 3x)^{3/2}}dx
Evaluate ∫01x31+x2 dx\displaystyle \int_{0}^{1} \frac{x^3}{\sqrt{1+x^2}} \,dx
Evaluate the definite integral ∫0π/3ln⁡(cos⁡x)tan⁡x dx{\displaystyle \int_{0}^{\pi/3}}\ln{(\cos x)}\tan x \space dx.
Evaluate∫ln2ln3ex(ex−1)3 dx\text{Evaluate} \displaystyle \int_{ln2}^{ln3} e^x(e^x-1)^3 \space dx
Evaluate ∫ex∗3exdx\displaystyle \int e^x*3^{e^x}dx
Practice: Indefinite with Inverse Trig
Q.\textbf{Q.} Compute the following integral: ∫arcsinx1−x2 dx\displaystyle \int\frac{\text{arcsin}{x}}{\sqrt{1-x^2}}\,\text{d}x
final114
Evaluate the integral: ∫e2x1+e2xdx\displaystyle\int_{ }^{ }\frac{e^{2x}}{1+e^{2x}}dx
Evaluate the integral ∫cos⁡(x+1)x+1dx\displaystyle \int\frac{\cos(\sqrt{x+1})}{\sqrt{x+1}}dx. Use C for the constant of integration.
Evaluate ∫1e(ln⁡x−5)4x dx\displaystyle\int_1^e\frac{\left(\ln x-5\right)^4}{x}\ dx
Find ∫cot⁡x ln⁡(sin⁡x)dx\displaystyle\int_{ }^{ }\cot x\ \ln\left(\sin x\right)dx
Suppose f(x)f(x) is continuous and let ∫07f(x)dx=7\displaystyle\int_0^7f\left(x\right)dx=7 and ∫710f(x)dx=10 \displaystyle\int_7^{10}f\left(x\right)dx=10 , find 5∫02f(5x)dx\displaystyle5\int_0^2f\left(5x\right)dx.
Integration by Substitution: Definite with Exponential
Evaluate the integral: ∫01x3 3x4+2 dx\displaystyle\int_0^1x^3 \ 3^{x^4+2}\ dx
Integration by Substitution: Indefinite with Trig
Find the following integral: ∫1x2sin⁡(1x)cos⁡(1x) dx\displaystyle\int\frac{1}{x^2}\sin\left(\frac{1}{x}\right)\cos\left(\frac{1}{x}\right) \ dx.
Integration by Substitution: Indefinite with Square Root
Compute the following integral: ∫3x+2 dx\displaystyle \int \sqrt{3x+2}\,\text{d}x
Integration by Substitution: Indefinite with Fraction
Compute the following integral: ∫11+2x2 dx\displaystyle \int \frac{1}{1+2x^2} \ \text{d}x
Practice: Definite with Roots
Q.\textbf{Q.} Evaluate ∫14xx2+1 dx\displaystyle \int_{1}^{4}x\sqrt{x^2+1}\,\text{d}x
Practice: Indefinite with Root
Q:\textbf{Q:} Evaluate ∫2xx2−1dx\displaystyle\int_{ }^{ }\frac{2x}{\sqrt{x^2-1}}dx
Integration by Substitution: Powers and Root
Compute the following integral: ∫ x3x2+1 dx\displaystyle \int\,x^3\sqrt{x^2+1} \ \text{d}x
Integration by Substitution: Indefinite with Exponentials
Compute the following integral: ∫ez+ez dz\displaystyle \int e^{z+e^z}\,\text{d}z
Integration by Substitution: Indefinite with Root
Compute the following integral: ∫31−4x2 dx\displaystyle \int\frac{3}{\sqrt{1-4x^2}}\,\text{d}x
Integration by Substitution: Indefinite with Power
Compute the following integral: ∫(x−6)(x+6)1/3 dx\displaystyle \int (x-6)(x+6)^{1/3}\,\text{d}x
Integration by Substitution: Definite with Exponential
Compute the following integral: ∫13xe2x25 dx\displaystyle \int_{1}^{3}\frac{xe^{2x^2}}{5}\,\text{d}x
Evaluate the integral ∫01x3⋅3x4+2 dx\displaystyle\int_0^1x^3\cdotp3^{x^4+2}\ dx.
Integration by Substitution
Evaluate
∫−ππ/2xcos⁡(2x2+1) dx\int_{-\pi}^{\pi/2} x \cos(2x^2+1)\ dx
Integration by Substitution
Evaluate
∫1e(ln⁡x−5)2xdx\displaystyle\int_1^e\dfrac{\left(\ln x-5\right)^2}{x}dx
Practice Question: Integration by Substitution
Find
∫2x32x2+1dx\displaystyle\int_{ }^{ }\frac{2x^3}{2x^2+1}dx
Practice Question: Integration by Substitution
Find
∫cot⁡x ln⁡(sin⁡x)dx\displaystyle\int\cot x\ \ln\left(\sin x\right)dx
Evaluate the integral ∫1arcsin⁡x1−x2dx\displaystyle\int_{ }^{ }\frac{1}{\arcsin x\sqrt{1-x^2}}dx. (If you need to input an inverse trigonometric function, the computer asks that you use the "arc" notation instead of "-1", i.e. arcsin⁡\arcsin instead of sin⁡−1\sin^{-1}.)
Find the indefinite integral ∫x1+x2dx\int\frac{x}{1+x^2}dx.
Evaluate the integral ∫x3(1+x4)14dx\displaystyle \int_{ }^{ }x^3(1+x^4)^{\frac{1}{4}}dx.
Find the definite integral ∫14exxdx\displaystyle\int_1^4\frac{e^{\sqrt{x}}}{\sqrt{x}}dx.
Special u-substitution
Evaluate ∫xx−53  ⁣dx\int\frac{x}{\sqrt[3]{x-5}}\de{x}.
Combination of techniques: Integration by substitution
Determine ∫1(1+x)2  ⁣dx\int\frac{1}{(1+\sqrt{x})^2}\de{x}.
Evaluate the integral ∫1ecos⁡(ln⁡x)xdx\displaystyle\int_1^e\frac{\cos\left(\ln x\right)}{x} \text{d}x.
Evaluate the integral ∫1arcsin⁡x1−x2dx\displaystyle\int_{ }^{ }\frac{1}{\arcsin x\sqrt{1-x^2}}dx.
Find the value of the definite integral ∫14exxdx\displaystyle \int_1^4\frac{e^{\sqrt{x}}}{\sqrt{x}}dx.
Evaluate the integral ∫1ecos⁡(ln⁡x)xdx\displaystyle\int_1^e\frac{\cos\left(\ln x\right)}{x}dx.
Evaluate the integral ∫sin⁡2x1+sin⁡2xdx\displaystyle\int_{ }^{ }\frac{\sin2x}{1+\sin^2x}dx.
Evaluate the integral ∫cos⁡x sec⁡2(sin⁡x)dx\displaystyle\int_{ }^{ }\cos x\ \sec^2\left(\sin x\right)dx.
Find ∫2x32x2+1dx\displaystyle\int_{ }^{ }\frac{2x^3}{2x^2+1}dx
Find ∫3x(x2−6)dx\displaystyle\int_{ }^{ }\frac{3x}{\sqrt{\left(x^2-6\right)}}dx.
Evaluate ∫−102x23−3x3dx\int_{-1}^0\frac{2x^2}{3-3x^3}dx.
Find ∫12exxdx\int_{ }^{ }\frac{1}{2e^{\sqrt{x}}\sqrt{x}}dx
Integration by Substitution
Find the u expression needed to integrate ∫sec⁡2x tan⁡2x dx\int_{ }^{ }\sec^2x\ \tan^2x\ dx.
Integration by Substitution
Find the u expression needed to integrate ∫sec⁡2xtan⁡x dx\int_{ }^{ }\sec^2x\tan x\ dx.
Integration by Substitution
Find ∫5x2(x3−1)6dx\int_{ }^{ }5x^2\left(x^3-1\right)^6dx.
Integration by Substitution
Evaluate the integral ∫1e2ln⁡xxdx=\int_1^e\frac{2\ln x}{x}dx=
Integration by Substitution
If ∫25f(2x) dx=6\int_2^5f\left(2x\right)\ dx=6 and ∫210f(x)=20\int_2^{10}f\left(x\right)=20, find ∫24f(x) dx\int_2^4f\left(x\right)\ dx.
Integration by Substitution
∫e1xx2dx\displaystyle \int_{ }^{ }\frac{e^{\frac{1}{x}}}{x^2}dx
Practice: Property of Definite Integral
Practice: Property of Definite Integral
Suppose f(x)f(x) is continuous and let ∫07f(x)dx=7\displaystyle \int_0^7f\left(x\right)dx=7 and ∫710f(x)dx=10\displaystyle \int_7^{10}f\left(x\right)dx=10.
Find 5∫02f(5x)dx\displaystyle 5\int_0^2f\left(5x\right)dx.
Practice: Integration by Substitution (with back-substitution)
Practice: Integration by Substitution (with back-substitution)
Find ∫2x32x2+1dx\int_{ }^{ }\frac{2x^3}{2x^2+1}dx.
Practice: Integration by Substitution
Evaluate ∫02(x−2)e(x2−4x)dx\int_0^2\left(x-2\right)e^{\left(x^2-4x\right)}dx.
Practice: Integration by Substitution (~F2018 Final Q36)
Practice: Integration by Substitution
Find ∫cot⁡x ln⁡(sin⁡x)dx\int_{ }^{ }\cot x\ \ln\left(\sin x\right)dx.
Practice: Integration by Substitution (~F2017 Final Q35)
Practice: Integration by Substitution
Evaluate ∫1e(ln⁡x−5)2xdx\int_1^e\frac{\left(\ln x-5\right)^2}{x}dx.
Practice: Property of Definite Integral
Practice: Property of Definite Integral
Suppose f(x) is continuous and let ∫07f(x)dx=7\int_0^7f\left(x\right)dx=7 and ∫710f(x)dx=10\int_7^{10}f\left(x\right)dx=10. Find 5∫02f(5x)dx5\int_0^2f\left(5x\right)dx
Evaluate ∫01x31+x2 dx\displaystyle \int_{0}^{1} \frac{x^3}{\sqrt{1+x^2}} \,dx