Integrate standard functions and use substitution, integration by parts and partial fractions to evaluate a wide range of integrals

What this dot point is asking

SEAB wants you to integrate the standard functions and apply the three main techniques - substitution, integration by parts, and partial fractions - to evaluate integrals that do not yield to direct integration.

The answer

Standard integrals

The reverse of the standard derivatives, including:

Integration by substitution

Substitution reverses the chain rule. Choose $u$ so that $\dfrac{du}{dx}$ appears (up to a constant) in the integrand, rewrite the whole integral in $u$ (including $dx$), integrate, then revert to $x$. For definite integrals, change the limits to $u$-values.

Integration by parts

Reversing the product rule:

Choose $u$ to be the part that simplifies when differentiated (logs and powers) and $dv$ the part that integrates easily. A common guide is the LIATE order (logarithmic, inverse, algebraic, trigonometric, exponential) for picking $u$.

Partial fractions

A proper rational function with a factorable denominator splits into simpler fractions, each integrable as a logarithm or power. For example $\dfrac{1}{(x-1)(x+2)}$ becomes $\dfrac{A}{x-1} + \dfrac{B}{x+2}$, and each term integrates to a logarithm.

Recognising which technique a question wants

Choosing the right method quickly is half the battle. Reach for substitution when the integrand contains a function and (a constant multiple of) its own derivative, as in $\int x e^{x^2}\,dx$. Reach for integration by parts when the integrand is a product of two unrelated functions, one of which simplifies on differentiating, such as $\int x\ln x\,dx$. Reach for partial fractions when you have a proper rational function with a factorable denominator. A quick scan for these signatures, "derivative present", "product to peel apart", or "rational function", tells you the technique before you commit pen to paper and avoids the dead ends that cost time in an exam.

The "parts twice" and recurring-integral trick

Some integrals need integration by parts applied twice, and occasionally the original integral reappears, which you then solve algebraically. For $\int e^x \cos x\,dx$, applying parts twice brings back a multiple of the original integral $I$; collecting gives an equation like $I = e^x(\sin x + \cos x) - I$, so $2I = e^x(\sin x + \cos x)$ and $I = \tfrac{1}{2}e^x(\sin x + \cos x) + C$. Recognising that the integral has cycled back to itself, and solving for it as an unknown, is an elegant H2-level technique worth having ready for products of exponentials with sine or cosine.

Examples in context

Example 1. Total accumulated quantity. Integrating a rate of flow $r(t) = 3t e^{-t^2}$ over time uses the substitution $u = -t^2$ to recover the total volume delivered, the workhorse of accumulation problems.

Example 2. Logarithmic growth integral. Computing the work done against a resistance proportional to $\frac{1}{x}$ leads to a $\ln|x|$ integral, which is why logarithms appear naturally in energy and entropy calculations.

Try this

Q1. Find $\displaystyle\int (2x + 1)^4\,dx$. [2 marks]

• Cue. Substitution $u = 2x + 1$: $\frac{(2x+1)^5}{10} + C$.

Q2. Find $\displaystyle\int x e^x\,dx$. [3 marks]

• Cue. By parts with $u = x$, $dv = e^x dx$: $x e^x - e^x + C$.

Q3. Express $\dfrac{1}{x(x+1)}$ in partial fractions. [2 marks]

• Cue. $\dfrac{1}{x} - \dfrac{1}{x+1}$.