U-Substitution Method

The chain rule in reverse - substitution technique for integration

🔗 U-Substitution Method

What is U-Substitution?

U-substitution is the reverse of the chain rule for derivatives. It's the most important integration technique you'll learn!

💡 Key Idea: If an integral has the form "function times its derivative," use u-substitution!


The Chain Rule (Review)

Remember the chain rule for derivatives:

ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

Example: ddx[(x2+1)5]=5(x2+1)42x\frac{d}{dx}[(x^2+1)^5] = 5(x^2+1)^4 \cdot 2x


Reversing the Chain Rule

If we integrate both sides:

f(g(x))g(x)dx=f(g(x))+C\int f'(g(x)) \cdot g'(x)\,dx = f(g(x)) + C

This is the foundation of u-substitution!


The U-Substitution Process

Step-by-Step Method

Step 1: Identify uu (the "inside function")

  • Look for a composite function
  • Choose u=g(x)u = g(x)

Step 2: Find dudu

  • Calculate du=g(x)dxdu = g'(x)\,dx

Step 3: Rewrite the integral in terms of uu

  • Replace all xx terms with uu terms
  • Replace dxdx with dudu

Step 4: Integrate with respect to uu

Step 5: Substitute back to xx

  • Replace uu with the original expression

Step 6: Add +C+C


Example 1: Basic U-Substitution

Evaluate 2x(x2+1)5dx\int 2x(x^2+1)^5\,dx

Step 1: Choose uu

Let u=x2+1u = x^2 + 1 (the inside function)


Step 2: Find dudu

dudx=2x\frac{du}{dx} = 2x

du=2xdxdu = 2x\,dx


Step 3: Rewrite the integral

Notice that we have 2xdx2x\,dx in the original integral!

2x(x2+1)5dx=(x2+1)52xdx\int 2x(x^2+1)^5\,dx = \int (x^2+1)^5 \cdot 2x\,dx

=u5du= \int u^5\,du


Step 4: Integrate

=u66+C= \frac{u^6}{6} + C


Step 5: Substitute back

=(x2+1)66+C= \frac{(x^2+1)^6}{6} + C


Check (using chain rule): ddx[(x2+1)66]=6(x2+1)52x6=2x(x2+1)5\frac{d}{dx}\left[\frac{(x^2+1)^6}{6}\right] = \frac{6(x^2+1)^5 \cdot 2x}{6} = 2x(x^2+1)^5


Example 2: When You Need to Adjust

Evaluate x(x2+1)5dx\int x(x^2+1)^5\,dx

Step 1: Choose uu

u=x2+1u = x^2 + 1


Step 2: Find dudu

du=2xdxdu = 2x\,dx

So: xdx=12dux\,dx = \frac{1}{2}du


Step 3: Rewrite

x(x2+1)5dx=(x2+1)5xdx\int x(x^2+1)^5\,dx = \int (x^2+1)^5 \cdot x\,dx

=u512du= \int u^5 \cdot \frac{1}{2}du

=12u5du= \frac{1}{2}\int u^5\,du


Step 4: Integrate

=12u66+C=u612+C= \frac{1}{2} \cdot \frac{u^6}{6} + C = \frac{u^6}{12} + C


Step 5: Substitute back

=(x2+1)612+C= \frac{(x^2+1)^6}{12} + C


Pattern Recognition

Type 1: Exact Match

f(g(x))g(x)dx\int f'(g(x)) \cdot g'(x)\,dx

The derivative g(x)g'(x) appears exactly!

Example: 2x(x2+1)5dx\int 2x(x^2+1)^5\,dx

  • u=x2+1u = x^2+1, du=2xdxdu = 2x\,dx

Type 2: Constant Multiple Off

f(g(x))kg(x)dx\int f'(g(x)) \cdot k \cdot g'(x)\,dx

The derivative is off by a constant factor.

Example: x(x2+1)5dx\int x(x^2+1)^5\,dx

  • u=x2+1u = x^2+1, du=2xdxdu = 2x\,dx
  • Have xdx=12dux\,dx = \frac{1}{2}du

Type 3: Won't Work

If after substitution you still have xx terms mixed with uu, u-substitution won't work (try different uu or different method).


Trigonometric Examples

Example 3: Sine and Cosine

Evaluate sinxcos4xdx\int \sin x \cos^4 x\,dx

Step 1: Choose uu

u=cosxu = \cos x (we have its derivative sinx\sin x in the integral!)


Step 2: Find dudu

dudx=sinx\frac{du}{dx} = -\sin x

du=sinxdxdu = -\sin x\,dx

sinxdx=du\sin x\,dx = -du


Step 3: Rewrite

sinxcos4xdx=cos4xsinxdx\int \sin x \cos^4 x\,dx = \int \cos^4 x \cdot \sin x\,dx

=u4(du)=u4du= \int u^4 \cdot (-du) = -\int u^4\,du


Step 4: Integrate

=u55+C= -\frac{u^5}{5} + C


Step 5: Substitute back

=cos5x5+C= -\frac{\cos^5 x}{5} + C


Exponential Examples

Example 4: Exponential with Chain Rule

Evaluate xex2dx\int xe^{x^2}\,dx

Step 1: Choose uu

u=x2u = x^2 (the exponent)


Step 2: Find dudu

du=2xdxdu = 2x\,dx

xdx=12dux\,dx = \frac{1}{2}du


Step 3: Rewrite

xex2dx=ex2xdx\int xe^{x^2}\,dx = \int e^{x^2} \cdot x\,dx

=eu12du=12eudu= \int e^u \cdot \frac{1}{2}du = \frac{1}{2}\int e^u\,du


Step 4: Integrate

=12eu+C= \frac{1}{2}e^u + C


Step 5: Substitute back

=12ex2+C= \frac{1}{2}e^{x^2} + C


Example 5: Requiring Algebraic Manipulation

Evaluate xx2+1dx\int \frac{x}{x^2+1}\,dx

Step 1: Choose uu

u=x2+1u = x^2 + 1 (the denominator)


Step 2: Find dudu

du=2xdxdu = 2x\,dx

xdx=12dux\,dx = \frac{1}{2}du


Step 3: Rewrite

xx2+1dx=1x2+1xdx\int \frac{x}{x^2+1}\,dx = \int \frac{1}{x^2+1} \cdot x\,dx

=1u12du=121udu= \int \frac{1}{u} \cdot \frac{1}{2}du = \frac{1}{2}\int \frac{1}{u}\,du


Step 4: Integrate

=12lnu+C= \frac{1}{2}\ln|u| + C


Step 5: Substitute back

=12lnx2+1+C= \frac{1}{2}\ln|x^2+1| + C

Since x2+1>0x^2+1 > 0 always, we can write:

=12ln(x2+1)+C= \frac{1}{2}\ln(x^2+1) + C


When to Use U-Substitution

Good Candidates

Look for these patterns:

  1. Composite function with its derivative present

    • (x2+1)52xdx\int (x^2+1)^5 \cdot 2x\,dx
  2. Fraction where numerator is derivative of denominator

    • 2xx2+1dx\int \frac{2x}{x^2+1}\,dx
  3. Trig functions with derivatives

    • tanxdx=sinxcosxdx\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx (let u=cosxu = \cos x)
  4. Exponentials with derivatives of exponent

    • xex2dx\int xe^{x^2}\,dx

Common U-Choices

| Integral Type | Common Choice for uu | |---------------|----------------------| | (ax+b)n(ax+b)^n | u=ax+bu = ax+b | | f(x)[f(x)]nf'(x) \cdot [f(x)]^n | u=f(x)u = f(x) | | f(x)f(x)\frac{f'(x)}{f(x)} | u=f(x)u = f(x) (gives ln\ln ) | | ef(x)f(x)e^{f(x)} \cdot f'(x) | u=f(x)u = f(x) | | sinnxcosx\sin^n x \cos x | u=sinxu = \sin x | | cosnxsinx\cos^n x \sin x | u=cosxu = \cos x |


⚠️ Common Mistakes

Mistake 1: Choosing Wrong uu

WRONG: For x(x2+1)5dx\int x(x^2+1)^5\,dx, choosing u=xu = x

RIGHT: Choose u=x2+1u = x^2+1 (the composite function)


Mistake 2: Forgetting to Adjust Constants

WRONG: u=x2+1u = x^2+1, du=2xdxdu = 2x\,dx, and writing u5du\int u^5\,du for x(x2+1)5dx\int x(x^2+1)^5\,dx

RIGHT: Need 12\frac{1}{2} factor! u512du\int u^5 \cdot \frac{1}{2}du


Mistake 3: Not Substituting Everything

After substitution, you should have only uu and dudu - no xx remaining!

If xx remains, choose different uu or method won't work.


Mistake 4: Forgetting to Substitute Back

WRONG: Final answer u612+C\frac{u^6}{12} + C

RIGHT: Final answer (x2+1)612+C\frac{(x^2+1)^6}{12} + C

Always convert back to the original variable!


Mistake 5: Wrong Sign

When du=sinxdxdu = -\sin x\,dx, we have sinxdx=du\sin x\,dx = -du (negative!)

Don't forget the negative sign!


U-Substitution Checklist

Choose uu = inner function or complicated part

Find dudu and solve for needed differential

Rewrite integral completely in terms of uu

Check: No xx should remain after substitution

Integrate using basic formulas

Substitute back to original variable

Add +C+C

Check by differentiating (if time permits)


Practice Tips

Tip 1: Look for the Derivative

Ask: "What function's derivative do I see here?"

Tip 2: Try the "Inside" First

For composite functions like (x2+1)5(x^2+1)^5, try u=u = inside function first.

Tip 3: Don't Force It

If after substitution you still have messy xx and uu mixed, try different uu or different method.

Tip 4: Practice Pattern Recognition

The more you practice, the faster you'll recognize which uu to choose!


Advanced Example

Evaluate lnxxdx\int \frac{\ln x}{x}\,dx

Solution:

Let u=lnxu = \ln x

du=1xdxdu = \frac{1}{x}dx

lnxxdx=udu=u22+C=(lnx)22+C\int \frac{\ln x}{x}\,dx = \int u\,du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C


📝 Practice Strategy

  1. Identify the composite function or complicated part
  2. Look for its derivative somewhere in the integral
  3. Set uu equal to the inner function
  4. Calculate dudu and solve for needed form
  5. Substitute completely - no xx should remain!
  6. Integrate the simpler uu integral
  7. Substitute back to xx
  8. Always add +C+C
  9. Check your answer by differentiating

📚 Practice Problems

1Problem 1medium

Question:

Evaluate 2x(x2+1)5dx\int 2x(x^2 + 1)^5 \, dx using uu-substitution.

💡 Show Solution

Solution:

Let u=x2+1u = x^2 + 1

Then du=2xdxdu = 2x \, dx

Notice that 2xdx2x \, dx appears in the integral!

Substitute:

2x(x2+1)5dx=u5du\int 2x(x^2 + 1)^5 \, dx = \int u^5 \, du

Integrate:

=u66+C= \frac{u^6}{6} + C

Substitute back:

=(x2+1)66+C= \frac{(x^2 + 1)^6}{6} + C

2Problem 2medium

Question:

Evaluate (3x2+2)(x3+2x)7dx\int (3x^2+2)(x^3+2x)^7\,dx using u-substitution.

💡 Show Solution

Step 1: Identify uu

Let u=x3+2xu = x^3 + 2x (the inside function that's raised to the 7th power)


Step 2: Find dudu

dudx=3x2+2\frac{du}{dx} = 3x^2 + 2

du=(3x2+2)dxdu = (3x^2 + 2)\,dx


Step 3: Notice the pattern

We have exactly (3x2+2)dx(3x^2+2)\,dx in the integral! Perfect match!

(3x2+2)(x3+2x)7dx=(x3+2x)7(3x2+2)dx\int (3x^2+2)(x^3+2x)^7\,dx = \int (x^3+2x)^7 \cdot (3x^2+2)\,dx


Step 4: Substitute

=u7du= \int u^7\,du


Step 5: Integrate

=u88+C= \frac{u^8}{8} + C


Step 6: Substitute back

=(x3+2x)88+C= \frac{(x^3+2x)^8}{8} + C


Check: ddx[(x3+2x)88]=8(x3+2x)7(3x2+2)8=(3x2+2)(x3+2x)7\frac{d}{dx}\left[\frac{(x^3+2x)^8}{8}\right] = \frac{8(x^3+2x)^7(3x^2+2)}{8} = (3x^2+2)(x^3+2x)^7

Answer: (x3+2x)88+C\frac{(x^3+2x)^8}{8} + C

3Problem 3medium

Question:

Evaluate 2x(x2+1)5dx\int 2x(x^2 + 1)^5 \, dx using uu-substitution.

💡 Show Solution

Solution:

Let u=x2+1u = x^2 + 1

Then du=2xdxdu = 2x \, dx

Notice that 2xdx2x \, dx appears in the integral!

Substitute:

2x(x2+1)5dx=u5du\int 2x(x^2 + 1)^5 \, dx = \int u^5 \, du

Integrate:

=u66+C= \frac{u^6}{6} + C

Substitute back:

=(x2+1)66+C= \frac{(x^2 + 1)^6}{6} + C

4Problem 4hard

Question:

Evaluate 01xx2+1dx\int_0^1 \frac{x}{\sqrt{x^2 + 1}} \, dx.

💡 Show Solution

Solution:

Let u=x2+1u = x^2 + 1, then du=2xdxdu = 2x \, dx, so xdx=12dux \, dx = \frac{1}{2} du

Change limits:

  • When x=0x = 0: u=02+1=1u = 0^2 + 1 = 1
  • When x=1x = 1: u=12+1=2u = 1^2 + 1 = 2

Substitute:

01xx2+1dx=121u12du\int_0^1 \frac{x}{\sqrt{x^2 + 1}} \, dx = \int_1^2 \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du

=1212u1/2du= \frac{1}{2} \int_1^2 u^{-1/2} \, du

=12[u1/21/2]12= \frac{1}{2} \left[\frac{u^{1/2}}{1/2}\right]_1^2

=12[2u1/2]12= \frac{1}{2} \left[2u^{1/2}\right]_1^2

=[u1/2]12= \left[u^{1/2}\right]_1^2

=21=21= \sqrt{2} - \sqrt{1} = \sqrt{2} - 1

5Problem 5hard

Question:

Evaluate 01xx2+1dx\int_0^1 \frac{x}{\sqrt{x^2 + 1}} \, dx.

💡 Show Solution

Solution:

Let u=x2+1u = x^2 + 1, then du=2xdxdu = 2x \, dx, so xdx=12dux \, dx = \frac{1}{2} du

Change limits:

  • When x=0x = 0: u=02+1=1u = 0^2 + 1 = 1
  • When x=1x = 1: u=12+1=2u = 1^2 + 1 = 2

Substitute:

01xx2+1dx=121u12du\int_0^1 \frac{x}{\sqrt{x^2 + 1}} \, dx = \int_1^2 \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du

=1212u1/2du= \frac{1}{2} \int_1^2 u^{-1/2} \, du

=12[u1/21/2]12= \frac{1}{2} \left[\frac{u^{1/2}}{1/2}\right]_1^2

=12[2u1/2]12= \frac{1}{2} \left[2u^{1/2}\right]_1^2

=[u1/2]12= \left[u^{1/2}\right]_1^2

=21=21= \sqrt{2} - \sqrt{1} = \sqrt{2} - 1

6Problem 6medium

Question:

Evaluate sin3xcosxdx\int \sin^3 x \cos x\,dx.

💡 Show Solution

Step 1: Choose uu

Let u=sinxu = \sin x (we have its derivative cosx\cos x in the integral)


Step 2: Find dudu

dudx=cosx\frac{du}{dx} = \cos x

du=cosxdxdu = \cos x\,dx


Step 3: Rewrite the integral

sin3xcosxdx=(sinx)3cosxdx\int \sin^3 x \cos x\,dx = \int (\sin x)^3 \cdot \cos x\,dx

=u3du= \int u^3\,du


Step 4: Integrate

=u44+C= \frac{u^4}{4} + C


Step 5: Substitute back

=sin4x4+C= \frac{\sin^4 x}{4} + C


Check: ddx[sin4x4]=4sin3xcosx4=sin3xcosx\frac{d}{dx}\left[\frac{\sin^4 x}{4}\right] = \frac{4\sin^3 x \cdot \cos x}{4} = \sin^3 x \cos x

Answer: sin4x4+C\frac{\sin^4 x}{4} + C

7Problem 7hard

Question:

Evaluate exxdx\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx.

💡 Show Solution

Step 1: Choose uu

Let u=x=x1/2u = \sqrt{x} = x^{1/2} (the exponent in exe^{\sqrt{x}})


Step 2: Find dudu

dudx=12x1/2=12x\frac{du}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}

du=12xdxdu = \frac{1}{2\sqrt{x}}\,dx

Solve for dxx\frac{dx}{\sqrt{x}}: 2du=dxx2\,du = \frac{dx}{\sqrt{x}}


Step 3: Rewrite the integral

exxdx=ex1xdx\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx = \int e^{\sqrt{x}} \cdot \frac{1}{\sqrt{x}}\,dx

=eu2du=2eudu= \int e^u \cdot 2\,du = 2\int e^u\,du


Step 4: Integrate

=2eu+C= 2e^u + C


Step 5: Substitute back

=2ex+C= 2e^{\sqrt{x}} + C


Check: ddx[2ex]=2ex12x=exx\frac{d}{dx}[2e^{\sqrt{x}}] = 2e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}}}{\sqrt{x}}

Answer: 2ex+C2e^{\sqrt{x}} + C