Methods of Integration

7.3 Methods of Integration

The method of inspection works for simple functions but quickly becomes impractical. To handle a wider variety of integrands, we use systematic techniques that transform unfamiliar integrals into standard forms. Three methods form the backbone:

1. Integration by Substitution
2. Integration using Partial Fractions
3. Integration by Parts

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1. Integration by Substitution

This method reverses the chain rule. If $f(g(x))$ has derivative $f'(g(x)) \cdot g'(x)$, substitution undoes this.

The Substitution Rule

Let $F$ be an antiderivative of $f$, so $\int f(u)\,du = F(u) + C$. If $u = g(x)$ is differentiable, then by the chain rule $\frac{d}{dx} F(g(x)) = f(g(x)) \cdot g'(x)$, and integrating both sides:

How to Apply Substitution

1. Choose $u = g(x)$ so that $g'(x)$ appears (up to a constant factor) in the integrand.
2. Compute $du = g'(x) \, dx$.
3. Rewrite the entire integral in terms of $u$ and $du$; all $x$ terms must be eliminated.
4. Integrate with respect to $u$.
5. Substitute back $u = g(x)$.

Substitution for Definite Integrals

For $\int_a^b f(g(x)) \cdot g'(x) \, dx$ with $u = g(x)$, either (i) find the indefinite integral in terms of $x$ and evaluate at $a$ and $b$, or (ii) change the limits to $u$-values — when $x = a$, $u = g(a)$; when $x = b$, $u = g(b)$:

The second method avoids substituting back.

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2. Integration using Partial Fractions

Many rational functions — quotients of polynomials — can be integrated by decomposing them into simpler fractions whose integrals are known.

Proper and Improper Rational Functions

$\frac{P(x)}{Q(x)}$ is proper if $\deg P < \deg Q$. If improper ($\deg P \geq \deg Q$), first perform polynomial long division:

where $\frac{R(x)}{Q(x)}$ is proper.

Partial Fraction Decomposition

The decomposition depends on how $Q(x)$ factorises. Four cases:

Case 1 — Distinct Linear Factors. If $Q(x) = (a_1x + b_1)(a_2x + b_2) \cdots (a_nx + b_n)$ with all factors distinct:

Case 2 — Repeated Linear Factors. A factor $(ax + b)^k$ contributes $k$ terms:

Case 3 — Distinct Irreducible Quadratic Factors. An irreducible $ax^2 + bx + c$ (discriminant $b^2 - 4ac < 0$) contributes:

Case 4 — Repeated Irreducible Quadratic Factors. A factor $(ax^2 + bx + c)^k$ contributes $k$ terms:

Finding the Constants

Multiply both sides by $Q(x)$ to clear denominators, then either substitute convenient values of $x$ (especially the roots of linear factors), or equate coefficients of like powers of $x$ and solve the resulting system.

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3. Integration by Parts

This method reverses the product rule. If $u$ and $v$ are differentiable functions of $x$, then $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$. Integrating gives $uv = \int u \, dv + \int v \, du$, so:

Choosing $u$ and $dv$

Choose $u$ by the ILATE priority order, and $dv$ as the remaining part (including $dx$):

Repeated Integration by Parts

Sometimes one application of integration by parts is not enough.

›Proof

Example: Find $\int x^2 e^x \, dx$.

Let $u = x^2$, $dv = e^x \, dx$. Then $du = 2x \, dx$, $v = e^x$.

$\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx = x^2 e^x - 2 \int x e^x \, dx$

Applying integration by parts to $\int x e^x \, dx = x e^x - e^x + C_1$:

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