Integration Techniques: A Practical Guide to Finding Antiderivatives

Finding antiderivatives is one of the central skills in calculus. While basic integrals can be solved with the power rule, many real-world problems require more sophisticated techniques. This guide walks through the most important integration methods: substitution, integration by parts, partial fractions, and trigonometric substitution.

Each technique is explained with the underlying logic, a clear formula, and fully worked examples so you can apply them confidently.

1. u-Substitution (The Chain Rule in Reverse)

Substitution is the most frequently used integration technique. It works when you can identify an inner function whose derivative also appears in the integrand. The formula is:

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x)

Example 1: Basic Substitution

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

Let $u = x^2$ , so $du = 2x \, dx$ . The integral becomes:

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

Example 2: Adjusting the Constant

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

Let $u = 3x^2$ , so $du = 6x \, dx$ , meaning $x \, dx = \frac{du}{6}$ .

\int x e^{3x^2} \, dx = \frac{1}{6} \int e^u \, du = \frac{1}{6} e^{3x^2} + C

2. Integration by Parts

Integration by parts is the product rule in reverse. Use it when the integrand is a product of two functions where one becomes simpler when differentiated. The formula is:

\int u \, dv = uv - \int v \, du

A common mnemonic for choosing $u$ is LIATE: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Pick $u$ from whichever type appears first in this list.

Example 3: Integration by Parts

Evaluate $\int x e^x \, dx$ .

Choose $u = x$ (algebraic) and $dv = e^x \, dx$ , so $du = dx$ and $v = e^x$ .

\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C = e^x(x - 1) + C

Example 4: Repeated Integration by Parts

Evaluate $\int x^2 \sin x \, dx$ .

First pass: $u = x^2$ , $dv = \sin x \, dx$ , giving $du = 2x \, dx$ , $v = -\cos x$ .

\int x^2 \sin x \, dx = -x^2 \cos x + 2\int x \cos x \, dx

Second pass on the remaining integral: $u = x$ , $dv = \cos x \, dx$ .

= -x^2 \cos x + 2\left(x \sin x - \int \sin x \, dx\right)

= -x^2 \cos x + 2x \sin x + 2\cos x + C

3. Partial Fractions

Partial fraction decomposition breaks a complex rational function into simpler fractions that are easier to integrate individually. This technique applies when the integrand is a ratio of polynomials and the degree of the numerator is less than the degree of the denominator.

Example 5: Distinct Linear Factors

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

Decompose:

\frac{3x + 5}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3}

Multiplying both sides by $(x+1)(x+3)$ :

3x + 5 = A(x+3) + B(x+1)

Setting $x = -1$ : $2 = 2A$ , so $A = 1$ . Setting $x = -3$ : $-4 = -2B$ , so $B = 2$ .

\int \frac{3x+5}{(x+1)(x+3)} \, dx = \int \frac{1}{x+1} \, dx + \int \frac{2}{x+3} \, dx = \ln|x+1| + 2\ln|x+3| + C

4. Trigonometric Substitution

When the integrand contains expressions like $\sqrt{a^2 - x^2}$ , $\sqrt{a^2 + x^2}$ , or $\sqrt{x^2 - a^2}$ , a trigonometric substitution can eliminate the square root. The three standard substitutions are:

$\sqrt{a^2 - x^2}$ : let $x = a\sin\theta$
$\sqrt{a^2 + x^2}$ : let $x = a\tan\theta$
$\sqrt{x^2 - a^2}$ : let $x = a\sec\theta$

Example 6: Trig Substitution

Evaluate $\int \frac{dx}{\sqrt{4 - x^2}}$ .

Here $a = 2$ . Let $x = 2\sin\theta$ , so $dx = 2\cos\theta \, d\theta$ and $\sqrt{4 - x^2} = 2\cos\theta$ .

\int \frac{2\cos\theta \, d\theta}{2\cos\theta} = \int d\theta = \theta + C = \arcsin\!\left(\frac{x}{2}\right) + C

Choosing the Right Technique

A quick decision tree:

Is there an inner function whose derivative also appears? Use substitution.
Is the integrand a product of two different function types? Use integration by parts (LIATE rule).
Is it a rational function (polynomial over polynomial)? Use partial fractions.
Does it contain a square root of a quadratic? Use trigonometric substitution.

Common Mistakes

Forgetting $+ C$ . Every indefinite integral needs the constant of integration.
Not converting $dx$ in substitution. When you substitute $u = g(x)$ , you must also replace $dx$ with $\frac{du}{g'(x)}$ .
Choosing the wrong $u$ in by-parts. Follow the LIATE order. If the integral gets harder after one pass, you probably chose the wrong function.

Try It Yourself

Evaluate integrals instantly with our free Integral Calculator for symbolic antiderivatives, or our Definite Integral Calculator for numerical results over a specific interval.