u-Substitution

Part 1 of 7 — Basic u-Substitution

The Idea

u-Substitution is the reverse of the Chain Rule. When you see a composite function inside an integral, you substitute $u$ for the inner function.

The Method

1. Identify the inner function — call it $u$
2. Compute $du = u'(x)\,dx$
3. Rewrite the entire integral in terms of $u$ and $du$
4. Integrate in terms of $u$
5. Substitute back to $x$

Worked Example 1

$\int 2x \cos(x^2)\,dx$

Let $u = x^2$, so $du = 2x\,dx$.

$\int \cos(u)\,du = \sin(u) + C = \sin(x^2) + C$

Worked Example 2

$\int 3x^2 e^{x^3}\,dx$

Let $u = x^3$, so $du = 3x^2\,dx$.

$\int e^u\,du = e^u + C = e^{x^3} + C$

Worked Example 3

$\int \frac{5}{(5x+1)^3}\,dx$

Let $u = 5x + 1$, so $du = 5\,dx$, meaning $dx = \frac{du}{5}$.

$\int \frac{5}{u^3} \cdot \frac{du}{5} = \int u^{-3}\,du = \frac{u^{-2}}{-2} + C = -\frac{1}{2(5x+1)^2} + C$

Basic u-Substitution 🎯

Key Takeaways — Part 1

1. u-Substitution reverses the Chain Rule
2. Choose $u$ = inner function of a composition
3. Compute $du$, then rewrite everything in terms of $u$
4. If $du$ doesn't appear exactly, adjust with constants

u-Substitution

Part 2 of 7 — Adjusting for Missing Constants

When $du$ Doesn't Match Exactly

Often the coefficient doesn't match perfectly. You can multiply and divide by constants to fix this.

Worked Example

$\int x^2 e^{x^3}\,dx$

Let $u = x^3$, then $du = 3x^2\,dx$. We have $x^2\,dx$ but need $3x^2\,dx$.

$\int x^2 e^{x^3}\,dx = \frac{1}{3}\int 3x^2 e^{x^3}\,dx = \frac{1}{3}\int e^u\,du = \frac{e^{x^3}}{3} + C$

Common Patterns to Recognize

Pattern Recognition 🎯

Key Takeaways — Part 2

1. You can adjust by constant multiples — multiply and divide to match $du$
2. Recognize patterns: $\frac{f'}{f} \to \ln|f|$, power-of-trig times derivative-of-trig
3. You CANNOT move a variable ($x$) outside the integral — only constants

u-Substitution

Part 3 of 7 — u-Substitution with Definite Integrals

Two Approaches

Method 1: Change the limits (recommended) When you substitute $u = g(x)$, change the limits: if $x = a$ then $u = g(a)$, if $x = b$ then $u = g(b)$.

Method 2: Back-substitute Find the antiderivative in terms of $x$, then evaluate at the original limits.

Worked Example — Method 1

$\int_0^2 x(x^2+1)^3\,dx$

Let $u = x^2 + 1$, $du = 2x\,dx$, so $x\,dx = \frac{du}{2}$.

Change limits: $x=0 \Rightarrow u=1$, $x=2 \Rightarrow u=5$.

$\frac{1}{2}\int_1^5 u^3\,du = \frac{1}{2} \cdot \frac{u^4}{4}\bigg|_1^5 = \frac{1}{8}(625 - 1) = \frac{624}{8} = 78$

Definite Integrals with u-Sub 🎯

Key Takeaways — Part 3

1. For definite integrals, change the limits to $u$-values to avoid back-substituting
2. If $u = g(x)$: new lower = $g(a)$, new upper = $g(b)$
3. After changing limits, evaluate entirely in $u$ — never mix $u$ and $x$

u-Substitution

Part 4 of 7 — Trickier Substitutions

Exponential and Logarithmic Substitutions

$\int \frac{e^x}{1+e^x}\,dx$

Let $u = 1 + e^x$, $du = e^x\,dx$.

$\int \frac{du}{u} = \ln|u| + C = \ln(1 + e^x) + C$

Substitution with Square Roots

$\int \frac{x}{\sqrt{x+1}}\,dx$

Let $u = x + 1$, so $x = u - 1$ and $dx = du$.

$\int \frac{u-1}{u^{1/2}}\,du = \int (u^{1/2} - u^{-1/2})\,du = \frac{2}{3}u^{3/2} - 2u^{1/2} + C$

$= \frac{2}{3}(x+1)^{3/2} - 2\sqrt{x+1} + C$

Trickier u-Substitutions 🎯

Key Takeaways — Part 4

1. For $\frac{f'}{f}$ patterns, use $u = f$ to get $\ln|f| + C$
2. Sometimes you need to express $x$ in terms of $u$ (e.g., $x = u - 1$)
3. Trig powers: use $u = \sin x$ or $u = \tan x$ when the derivative appears

u-Substitution

Part 5 of 7 — Long Division and Completing the Square

When u-Sub Doesn't Work Directly

Some integrals need algebraic manipulation before substitution.

Long Division for Improper Fractions

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

Divide first: $\frac{x^2+1}{x+1} = x - 1 + \frac{2}{x+1}$

$\int \left(x - 1 + \frac{2}{x+1}\right)dx = \frac{x^2}{2} - x + 2\ln|x+1| + C$

Completing the Square

$\int \frac{1}{x^2 + 4x + 8}\,dx = \int \frac{1}{(x+2)^2 + 4}\,dx$

Let $u = x + 2$:

$\int \frac{1}{u^2 + 4}\,du = \frac{1}{2}\arctan\left(\frac{u}{2}\right) + C = \frac{1}{2}\arctan\left(\frac{x+2}{2}\right) + C$

Algebraic Manipulation + Integration 🎯

Key Takeaways — Part 5

1. Long division when the degree of numerator $\geq$ degree of denominator
2. Complete the square when you see $x^2 + bx + c$ in a denominator
3. After algebraic prep, standard techniques (u-sub, $\arctan$) apply

u-Substitution

Part 6 of 7 — Problem-Solving Workshop

Mixed problems combining all u-substitution techniques.

Mixed u-Sub Problems 🎯

More Challenging Problems 🎯

Workshop Complete!

You practiced u-sub with exponentials, logarithms, trig functions, and algebraic manipulation.

u-Substitution Review

Part 7 of 7 — Comprehensive Assessment

Quick Reference

Final Assessment 🎯

u-Substitution — Complete! ✅

You have mastered:

• ✅ Basic u-substitution
• ✅ Adjusting for missing constants
• ✅ Definite integrals with changed limits
• ✅ Trickier substitutions (exponential, log, trig)
• ✅ Long division and completing the square
• ✅ Pattern recognition strategies