U-Substitution

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Integration by Substitution (a.k.a. "u-Substitution)
It's like "reverse chain rule" for integration.

Identifying Clues
If you see a function wrapped up in another function, and its derivative shows up as well
If you see both a sin⁡\sinsin and a cos⁡\coscos
If you see a ln⁡x\ln xlnx and some xxx in the denominator
If you see a x\sqrt{x}x​ or 1x\frac{1}{x}x1​ wrapped inside another function
Strategy
1. Pick one part of the function and let that be uuu
u′u'u′ should appear elsewhere in the function
Pick uuu as the function wrapped inside another
Pick uuu as ln⁡x\ln xlnx
Pick uuu as the denominator
Sometimes we need to factor or complete the square before we do our u-substitution
2. Differentiate uuu → du=...dxdu=...dxdu=...dx

3. Solve for dxdxdx

4. Substitute uuu , replace dxdxdx
For definite integrals, we also want to replace the upper and lower integral bounds with u-bounds
If you still have x's remaining in your integral after substitution, go back to your u=...u=...u=... formula and do another substitution

5. Integrate the new function with respect to uuu

6. Subsitute the original xxx expression back in for uuu

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Example
Evaluate ∫2xx2−1dx\displaystyle\int_{ }^{ }\frac{2x}{\sqrt{x^2-1}}dx∫​x2−1​2x​dx

Let u=x2−1u=x^2-1u=x2−1
du=2x dxdu=2x\ dxdu=2x dx
dx=12xdudx=\frac{1}{2x}dudx=2x1​du
∫2xu(12xdu)\displaystyle \int_{ }^{ }\frac{2x}{\sqrt{u}}\left(\frac{1}{2x}du\right)∫​u​2x​(2x1​du)
=∫1u du=\displaystyle \int \frac{1}{\sqrt{u}}\ du=∫u​1​ du
=u1212+C=\frac{u^{\frac{1}{2}}}{\frac{1}{2}}+C=21​u21​​+C
=2u+C=2\sqrt{u}+C=2u​+C
2x2−1+C2\sqrt{x^2-1}+C2x2−1​+C

Practice: Integration by Substitution
Evaluate ∫e1xx2dx\displaystyle \int_{ }^{ }\frac{e^{\frac{1}{x}}}{x^2}dx∫​x2ex1​​dx

e1x+Ce^{\frac{1}{x}}+Cex1​+C

−e1x+C-e^{\frac{1}{x}}+C−ex1​+C

−e1xx2+C-\frac{e^{\frac{1}{x}}}{x^2}+C−x2ex1​​+C

−3e1xx3+C-\frac{3e^{\frac{1}{x}}}{x^3}+C−x33ex1​​+C

3e1xx3+C\frac{3e^{\frac{1}{x}}}{x^3}+Cx33ex1​​+C

I don't know

Example: Integration by Substitution
Evaluate ∫1x+x1/2dx\displaystyle \int_{ }^{ }\frac{1}{x+x^{1/2}}dx∫​x+x1/21​dx

2ln⁡∣x1/2+1∣+C2\ln|x^{1/2}+1|+C2ln∣x1/2+1∣+C

ln⁡∣x1/2+x∣+C\ln|x^{1/2}+x|+Cln∣x1/2+x∣+C

ln⁡∣x1/2+x∣x+1+C\frac{\ln|x^{1/2}+x|}{x+1}+Cx+1ln∣x1/2+x∣​+C

ln⁡∣x∣+2x+C\ln|x|+2\sqrt{x}+Cln∣x∣+2x​+C

ln⁡∣x∣−2x+C\ln\left|x\right|-\frac{2}{\sqrt{x}}+Cln∣x∣−x​2​+C

I don't know

Practice: Substitution with Definite Integral
Evaluate ∫1e(ln⁡x−5)2xdx\displaystyle 
\int_1^e\frac{\left(\ln x-5\right)^2}{x}dx∫1e​x(lnx−5)2​dx.

612\frac{61}{2}261​

613\frac{61}{3}361​

−63-63−63

−32e2+1252-\frac{32}{e^2}+\frac{125}{2}−e232​+2125​

−32e+1252-\frac{32}{e}+\frac{125}{2}−e32​+2125​

I don't know

Practice: Integration by Substitution (with back-substitution)
Find ∫2x32x2+1dx\int_{ }^{ }\frac{2x^3}{2x^2+1}dx∫​2x2+12x3​dx.

ln⁡∣x+1∣2x2+C\frac{\ln\left|x+1\right|}{2x^2}+C2x2ln∣x+1∣​+C

2x2+14+C\frac{2x^2+1}{4}+C42x2+1​+C

2x2+1−ln⁡∣2x2+1∣4+C\frac{2x^2+1-\ln\left|2x^2+1\right|}{4}+C42x2+1−ln∣2x2+1∣​+C

ln⁡∣2x2+1∣2x2+1+C\frac{\ln\left|2x^2+1\right|}{2x^2+1}+C2x2+1ln∣2x2+1∣​+C

None of the above

I don't know

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Short-Cut for U-Substitution
Instead of going through the entire process of integration by substitution (u-sub), there is a short-cut for the case where the argument is only changed by a linear term.

Examples
1. ∫cos⁡(2x) dx=\int_{ }^{ }\cos\left(2x\right)\ dx=∫​cos(2x) dx=

2. ∫e−3x dx=\int_{ }^{ }e^{-3x}\ dx=∫​e−3x dx=

3. ∫14x−7 dx=\int_{ }^{ }\frac{1}{4x-7}\ dx=∫​4x−71​ dx=

Wize University Calculus 2 Textbook > Integration

Popular Courses

Integration by Substitution (a.k.a. "u-Substitution)

Practice: Integration by Substitution

Example: Integration by Substitution

Practice: Substitution with Definite Integral

Practice: Integration by Substitution (with back-substitution)

Short-Cut for U-Substitution

Examples

Extra Practice

Practice: Integration by Substitution (~F2017 Final Q35)

Practice: Integration by Substitution (~2019 Test1 A10)

Practice Question: U-Sub