What does u stand for in u-substitution?

Nothing special: u is just a new letter for the inside function, chosen so the integral looks simpler in terms of u.

How do I know what to pick for u?

Pick an inside function whose derivative also appears in the integrand, up to a constant factor. If after substituting some x is left over, try a different u.

Do you need to change the limits in u-substitution?

For definite integrals, either change the limits to u-values and stay in u, or substitute back to x and use the original limits. Don't mix the two.

Why do I divide by a constant in u-substitution?

When du differs from what's in the integral by a number, such as du=2xdx when you have xdx, you divide by that number. You can only do this for constants, not for expressions with x.

How can I check my answer?

Differentiate it. By the chain rule you should get back exactly the original integrand.

U-Substitution: How to Choose u, With Worked Examples

What is u-substitution?

The basic integration rules handle powers, $e^{x}$ , $\frac{1}{x}$ , $\sin x$ and $\cos x$ , but not something like $\int 2 x \cos (x^{2}) d x$ . The way in is to remember where such an integrand comes from. By the chain rule, $\frac{d}{d x} \sin (x^{2}) =$ $\cos (x^{2}) \cdot 2 x$ : an outside function's derivative, evaluated at the inside, times the inside's derivative. Our integrand has exactly that shape, so its antiderivative must be $\sin (x^{2}) + C$ .

U-substitution (also called integration by substitution) makes this reverse-chain-rule idea systematic. You give the inside function a new name, $u$ , which turns the integral into a simpler one in $u$ .

The substitution rule

u = g (x)

, then

\int f (g (x)) g^{'} (x) d x =

\int f (u) d u

Why it works

Let $F$ be an antiderivative of $f$ , so $F^{'} = f$ .
By the chain rule, $\frac{d}{d x} F (g (x)) =$ $f (g (x)) g^{'} (x)$ .
So $F (g (x)) + C$ is the antiderivative of the left side, and $F (u) + C$ with $u = g (x)$ is exactly the same thing.

How do you do u-substitution step by step?

Choose $u$ : usually the inside function (inside a power, a root, $e^{()}$ , $\sin ()$ , or the bottom of a fraction).
Find $d u$ : differentiate, $\frac{d u}{d x} = u^{'} (x)$ , and write it as $d u = u^{'} (x) d x$ .
Rewrite the whole integral in terms of $u$ and $d u$ . Every $x$ must disappear; adjust a constant factor if needed.
Integrate with respect to $u$ .
Substitute back: replace $u$ with the original expression in $x$ . Add $+ C$ .

Worked example: A power of an inside function

Find $\int x (x^{2} + 1)^{5} d x$ .

Choose: the inside of the 5th power, $u = x^{2} + 1$ .
Differentiate: $d u = 2 x d x$ . The integral has $x d x$ , not $2 x d x$ , so divide by $2$ : $x d x = \frac{1}{2} d u$ .
Rewrite: $\int (x^{2} + 1)^{5} x d x =$ $\int u^{5} \cdot \frac{1}{2} d u =$ $\frac{1}{2} \int u^{5} d u$ .
Integrate: $\frac{1}{2} \cdot \frac{u^{6}}{6} =$ $\frac{u^{6}}{12}$ .
Substitute back: $\frac{(x^{2} + 1)^{6}}{12} + C$ .
Check by differentiating: $\frac{6 (x^{2} + 1)^{5} \cdot 2 x}{12} =$ $x (x^{2} + 1)^{5}$ . ✓

Answer: $\frac{(x^{2} + 1)^{6}}{12} + C$

Worked example: A linear inside function

Find $\int \cos (3 x) d x$ .

$u = 3 x$ , so $d u = 3 d x$ and $d x = \frac{1}{3} d u$ .
$\int \cos u \cdot \frac{1}{3} d u =$ $\frac{1}{3} \sin u$ .
Substitute back: $\frac{1}{3} \sin (3 x) + C$ .

Answer: $\frac{1}{3} \sin (3 x) + C$

Tip

The pattern in the last example is worth remembering: when the inside is $a x + b$ , the integral is the usual one divided by $a$ . For example $\int e^{5 x} d x =$ $\frac{1}{5} e^{5 x} + C$ .

How do you choose u?

Look for a piece of the integrand whose derivative also appears (up to a constant number). These patterns cover most problems:

Integrand looks like	Choose $u =$	Example
(inside) $^{n}$ times the inside's derivative	the inside	$\int x^{2} (x^{3} + 5)^{4} d x$ : $u = x^{3} + 5$
$\sqrt{inside}$ times the inside's derivative	the inside	$\int \frac{x}{\sqrt{x^{2} + 9}} d x$ : $u = x^{2} + 9$
$e^{inside}$ or $\sin (inside)$	the exponent or the angle	$\int x e^{x^{2}} d x$ : $u = x^{2}$
a fraction whose top is the bottom's derivative	the bottom	$\int \frac{2 x}{x^{2} + 1} d x$ : $u = x^{2} + 1$
$\ln x$ together with $\frac{1}{x}$	$\ln x$	$\int \frac{\ln x}{x} d x$ : $u = \ln x$

Worked example: A logarithm

Find $\int \frac{\ln x}{x} d x$ .

$u = \ln x$ , so $d u = \frac{1}{x} d x$ , which is exactly the rest of the integrand.
$\int u d u =$ $\frac{u^{2}}{2}$ .
Substitute back: $\frac{(\ln x)^{2}}{2} + C$ .

Answer: $\frac{(\ln x)^{2}}{2} + C$

Worked example: The integral of tan x

Find $\int \tan x d x$ .

Rewrite: $\tan x =$ $\frac{\sin x}{\cos x}$ . The top is (minus) the derivative of the bottom.
$u = \cos x$ , so $d u = - \sin x d x$ , and $\sin x d x = - d u$ .
$\int \frac{- d u}{u} =$ $- \ln | u |$ .
Substitute back: $- \ln | \cos x | + C$ .

Answer: $- \ln | \cos x | + C$

Common mistake

Adjusting for a missing $x$ . You can fix a missing constant (like the $\frac{1}{2}$ in the first example), but not a missing variable. In $\int (x^{2} + 1)^{5} d x$ , there's no $x$ to go with $d u = 2 x d x$ , so this substitution doesn't work; multiply out instead.

How do you use u-substitution with definite integrals?

You can substitute back to $x$ and use the original limits, but it's usually quicker to change the limits: when $x = a$ , $u = g (a)$ , and when $x = b$ , $u = g (b)$ . Then finish entirely in $u$ .

Substitution in definite integrals

\int_{a}^{b} f (g (x)) g^{'} (x) d x =

\int_{g (a)}^{g (b)} f (u) d u

Worked example: Changing the limits

Evaluate $\int_{0}^{1} x^{2} (x^{3} +$ $1)^{4} d x$ .

$u = x^{3} + 1$ , so $d u = 3 x^{2} d x$ and $x^{2} d x = \frac{1}{3} d u$ .
New limits: $x = 0$ gives $u = 0^{3} + 1 = 1$ ; $x = 1$ gives $u = 1 + 1 = 2$ .
$\int_{1}^{2} u^{4} \cdot \frac{1}{3} d u =$ $\frac{1}{3} [\frac{u^{5}}{5}]_{1}^{2}$ .
$=$ $\frac{1}{3} \cdot \frac{2^{5} - 1^{5}}{5} =$ $\frac{1}{3} \cdot \frac{31}{5} =$ $\frac{31}{15}$ .

Answer: $\frac{31}{15}$

Common mistake

Using the old limits with the new variable. If you switch to $u$ , switch the limits too. Evaluating $\frac{u^{5}}{15}$ from $0$ to $1$ (the $x$ -limits) would give $\frac{1}{15}$ , which is wrong.

When doesn't u-substitution work?

If no choice of $u$ makes every $x$ disappear, substitution isn't the right tool. Products like $\int x e^{x} d x$ or $\int x \sin x d x$ have no inside-function pattern; they need integration by parts, which is usually taught in Calculus II. The course's Playbook helps you tell which method an integral needs.

U-substitution: how to choose u, with step-by-step examples