Revise Integration techniques: substitution, by parts and partial fractions for IB Mathematics: Analysis and Approaches (5.8)

Substitution is the chain rule reversed. It is the tool for a composite function — an inside function $g(x)$ sitting inside an outer function — when the derivative $g'(x)$ (or a constant multiple of it) also appears in the integrand. By naming the inside function $u$, you convert the whole integral into a simpler one in $u$.

Let $u = g(x)$, so $du = g'(x)\,dx$. Then $\displaystyle \int f(g(x))\,g'(x)\,dx = \int f(u)\,du$.

The single most important mechanical step is forming $du$ correctly: differentiate $u$, then treat $\frac{du}{dx}$ as a fraction to get $du = g'(x)\,dx$. Every $x$ and every $dx$ in the original integral must disappear, replaced entirely by $u$ and $du$. If any stray $x$ remains, your substitution is not yet complete.

Integration by parts is the product rule reversed. It is the tool for integrating a product of two different function types — for example an algebraic function times an exponential, or an algebraic function times a logarithm — where substitution does not apply because there is no inside-function-and-its-derivative pattern.

Integration by parts: $\displaystyle \int u\,dv = uv - \int v\,du$, equivalently $\displaystyle \int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$.

The whole method turns on choosing $u$ and $dv$ well. You differentiate $u$ (so you want it to get simpler) and integrate $dv$ (so you must be able to integrate it). The LIATE rule ranks function types to make this choice: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — choose $u$ as whichever type appears first in that list, and let $dv$ be the rest.

A rational function that you cannot integrate directly can often be rewritten as a sum of simpler fractions that you can. For a proper rational function whose denominator factorises into distinct linear factors, each factor contributes one simple fraction with a constant on top — and each of those integrates to a logarithm.

For distinct linear factors: $\dfrac{P(x)}{(ax+b)(cx+d)} = \dfrac{A}{ax+b} + \dfrac{B}{cx+d}$, with $\deg P < 2$.

Half the battle in Paper 1 is choosing the right method before you start. Work through the integrand's structure in order: it is faster to eliminate techniques than to force one that does not fit.

In Paper 1 an integration question is marked analytically: the credit is tied to specific lines of working, not just the final expression. Each mark is either a method mark (M) — for the correct approach, such as choosing a valid substitution or applying the by-parts formula — or an accuracy mark (A) — for correct working or the correct answer. An A mark is dependent on the M mark it follows, so a right answer built on a wrong method earns neither. Equivalent correct forms are accepted, and an indefinite integral must carry $+\,C$. Study how each mark below is earned by a specific line.

These three techniques are the backbone of everything that follows in HL calculus. Substitution and by parts reappear when you compute areas between curves and volumes of revolution, where the integrand is rarely standard. Partial fractions is the gateway to integrating more of the rational functions that arise in differential equations and in the logistic and related growth models. And the habit that matters most — reading an integrand's structure before choosing a method — is exactly the judgement Paper 1 rewards. Master the decision map, write the method on the page, and complex integrals become variations on a routine you already own.