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Wize AP Calculus (AB) Textbook > Integration & Accumulation of Change

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}