Integration by Substitution (u-Substitution)

Quick answer

Substitution is the chain rule run backward. When an integrand has the form f(g(x)) · g′(x) — a function of an inner function, times the inner function's derivative — set u = g(x), so du = g′(x) dx, and the integral becomes ∫ f(u) du. Integrate in u, then write the answer back in x. A missing constant factor can be fixed by multiplying and dividing. For a definite integral, change the limits to u-values too, and there is no need to go back to x.

What you'll learn

Recognize integrands of the form f(g(x)) · g′(x)
Carry out a substitution, including constant adjustments
Change the limits of a definite integral under substitution

The chain rule backward

The chain rule says $\tfrac{d}{dx}(x^2 + 1)^6 = 6(x^2 + 1)^5 \cdot 2x$ . So, reading it backward,

\int 12x(x^2 + 1)^5\,dx = (x^2 + 1)^6 + C

The integrand has an inner function, $x^2 + 1$ , and next to it that inner function’s derivative, $2x$ . Spotting that pattern is the whole skill; substitution is the bookkeeping that follows.

The method

Choose $u$ : usually the inner function.
Compute $du = g'(x)\,dx$ .
Rewrite the integral entirely in $u$ — no $x$ left over.
Integrate in $u$ .
Substitute back $u = g(x)$ .

For $\int 2x(x^2 + 1)^5\,dx$ : let $u = x^2 + 1$ , so $du = 2x\,dx$ . Then

\int 2x(x^2 + 1)^5\,dx = \int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2 + 1)^6}{6} + C

Why the substitution is legitimate

Suppose $F$ is an antiderivative of $f$ . By the chain rule,

\frac{d}{dx}F\big(g(x)\big) = F'\big(g(x)\big)\,g'(x) = f\big(g(x)\big)\,g'(x)

So $F(g(x))$ is an antiderivative of $f(g(x))\,g'(x)$ . Writing $u = g(x)$ and $du = g'(x)\,dx$ is shorthand for exactly this. Substitution does not invent a new rule; it is the chain rule, recognized inside an integral.

Definite integrals

With a definite integral, change the limits to $u$ -values as well:

\int_a^b f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du

Then evaluate in $u$ and stop; there is nothing to substitute back.

Worked examples

Common mistakes

Practice problems

Find $\int 3x^2(x^3 - 5)^4\,dx$ .

Answer

$\tfrac{(x^3 - 5)^5}{5} + C$

Full solution

$u = x^3 - 5$ , $du = 3x^2\,dx$ : $\int u^4\,du = \tfrac{u^5}{5} + C$ .
Find $\int \cos(5x)\,dx$ .

Answer

$\tfrac{1}{5}\sin(5x) + C$

Full solution

$u = 5x$ , $du = 5\,dx$ , so $dx = \tfrac{du}{5}$ : $\tfrac{1}{5}\int \cos u\,du$ .
Find $\int \tfrac{2x}{x^2 + 4}\,dx$ .

Answer

$\ln(x^2 + 4) + C$

Full solution

$u = x^2 + 4$ , $du = 2x\,dx$ : $\int \tfrac{du}{u} = \ln\lvert u \rvert + C$ . Since $x^2 + 4 > 0$ , the absolute value can go.
Find $\int e^{3x + 1}\,dx$ .

Answer

$\tfrac{1}{3}e^{3x + 1} + C$

Full solution

$u = 3x + 1$ , $du = 3\,dx$ : $\tfrac{1}{3}\int e^u\,du$ .
Find $\int x\sqrt{x^2 + 9}\,dx$ .

Answer

$\tfrac{1}{3}(x^2 + 9)^{3/2} + C$

Full solution

$u = x^2 + 9$ , $x\,dx = \tfrac{du}{2}$ : $\tfrac{1}{2}\int u^{1/2}\,du = \tfrac{1}{2} \cdot \tfrac{2}{3}u^{3/2}$ .
Find $\int \tfrac{\ln x}{x}\,dx$ .

Answer

$\tfrac{(\ln x)^2}{2} + C$

Full solution

$u = \ln x$ , $du = \tfrac{dx}{x}$ : $\int u\,du = \tfrac{u^2}{2} + C$ .
Evaluate $\int_0^2 x(x^2 + 1)^2\,dx$ .

Answer

$\tfrac{62}{3}$

Full solution

$u = x^2 + 1$ runs from $1$ to $5$ , and $x\,dx = \tfrac{du}{2}$ : $\tfrac{1}{2}\int_1^5 u^2\,du = \tfrac{1}{2} \cdot \tfrac{125 - 1}{3} = \tfrac{62}{3}$ .
Evaluate $\int_0^{\pi/2} \sin x \cos^2 x\,dx$ .

Answer

$\tfrac{1}{3}$

Full solution

$u = \cos x$ runs from $1$ to $0$ , and $\sin x\,dx = -du$ : $-\int_1^0 u^2\,du = \int_0^1 u^2\,du = \tfrac{1}{3}$ .
Find $\int \tfrac{1}{1 + 4x^2}\,dx$ .

Answer

$\tfrac{1}{2}\arctan(2x) + C$

Full solution

$u = 2x$ , $dx = \tfrac{du}{2}$ : $\tfrac{1}{2}\int \tfrac{du}{1 + u^2} = \tfrac{1}{2}\arctan u + C$ .
A student evaluates $\int_0^1 2x(x^2 + 1)\,dx$ with $u = x^2 + 1$ as $\int_0^1 u\,du = \tfrac{1}{2}$ . What went wrong?

Hint

What are the $u$ -values at $x = 0$ and $x = 1$ ?

Answer

The limits were not changed. They are $1$ to $2$ , and the integral is $\tfrac{3}{2}$ .

Full solution

At $x = 0$ , $u = 1$ ; at $x = 1$ , $u = 2$ . So the integral is $\int_1^2 u\,du = \tfrac{4 - 1}{2} = \tfrac{3}{2}$ .

Check directly: $\int_0^1 (2x^3 + 2x)\,dx = \tfrac{1}{2} + 1 = \tfrac{3}{2}$ .

Frequently asked questions

When should I use u-substitution?

When the integrand contains an inner function together with that inner function's derivative, possibly off by a constant factor — such as 2x and x² + 1 in 2x(x² + 1)⁵.

How do I choose u?

Usually u is the inner function: the expression inside a power, root, exponential or trig function, whose derivative also appears in the integrand.

What if the derivative is off by a constant?

Multiply and divide by that constant. For ∫ x e^(x²) dx, du = 2x dx, so x dx = du/2.

How do the limits change in a definite integral?

Replace each limit x = a and x = b with u = g(a) and u = g(b). Then evaluate in u, with no need to substitute back.

Why does substitution work?

By the chain rule, the derivative of F(g(x)) is f(g(x)) g′(x) when F′ = f. So F(g(x)) is an antiderivative of the integrand.

What to learn next

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