Further techniques and miscellaneous exercises
Master further integration techniques for Year 12 Specialist Mathematics in Queensland (QCAA). This is the capstone of the integration topic, where the challenge is no longer applying one rule but recognising which method fits a less obvious integrand — and combining more than one when it does not match a single form.
You will learn to read an integrand and choose between standard forms, the \(f'/f\) logarithm, substitution, integration by parts, partial fractions and trig identities, then combine them for harder cases — the problem-solving skill that ties the whole calculus course together.
Theory
Further integration techniques is the capstone of the Year 12 Specialist Mathematics (QCAA, Queensland) integration topic: given a less obvious integrand you must recognise which method fits and, when needed, combine more than one. This page draws the whole toolbox — standard forms, the \(f'/f\) logarithm, substitution, by parts, partial fractions and trig identities — into one decision framework.
By this point you have met several ways to integrate. Further integration techniques is not a new rule but a strategy: read the integrand, decide which tool applies, and apply it — sometimes more than one.
The toolbox is: standard forms (the inverse-trig integrals and \(\int\sec^2 x\,dx=\tan x+c\)); the \(f'/f\) logarithm form, \(\int\dfrac{f'(x)}{f(x)}\,dx=\ln|f(x)|+c\); substitution \(u=g(x)\) for \(f(g(x))\,g'(x)\); integration by parts for a product of unlike functions; partial fractions for a rational function with a factorable denominator; and trigonometric identities for even powers of \(\sin\) and \(\cos\).
Choosing well comes down to recognising a trigger pattern. A denominator of the shape \(\sqrt{a^2-x^2}\) or \(a^2+x^2\) points to an inverse-trig form; a numerator that is the derivative of the denominator points to \(\ln|f|\); a composite times its inner derivative points to substitution; a product with no inner-derivative factor points to parts; two distinct linear factors on the bottom point to partial fractions.
Harder integrands often need you to combine techniques — split a fraction into two pieces, simplify with an identity before integrating, or substitute and then integrate by parts. The skill this page builds is deciding the first move quickly and correctly.
The standard forms (with \(a>0\)) — recognise the shape of the integrand and read off the answer:
The logarithm form, when the top is the derivative of the bottom:
Integration by parts for a product, and the \(\sec^2\) standard form:
The power-reduction identities for even powers of sine and cosine:
How to choose a technique
- Read the integrand's shape: is it a known standard form (an inverse-trig denominator, \(\sec^2 x\)) or an even power of \(\sin\)/\(\cos\) needing an identity? If so, apply it directly.
- Check for \(f'/f\) or substitution: is the numerator the derivative of the denominator (use \(\ln|f|\)), or is there an inner function whose derivative is a factor (use \(u=g(x)\))?
- Otherwise classify by structure: a product of unlike functions uses integration by parts; a rational function with a factorable denominator uses partial fractions.
- Combine if it is not a single form: split the fraction, simplify with an identity, or substitute first — then integrate each resulting piece and add \(+c\).
The denominator is \(a^2+x^2\) with a constant on top, so this is the inverse-tan standard form \(\int\dfrac{1}{a^2+x^2}\,dx=\dfrac{1}{a}\arctan\dfrac{x}{a}+c\):
| \(a^2\) | \(=\) | \(49\) |
| \(a\) | \(=\) | \(7\) |
Apply the form with \(a=7\), keeping the \(\dfrac{1}{a}\) factor:
| \(\int \dfrac{1}{x^2+49}\,dx\) | \(=\) | \(\dfrac{1}{a}\arctan\dfrac{x}{a}+c\) |
| \(=\) | \(\dfrac{1}{7}\arctan\dfrac{x}{7}+c\) |
\(\displaystyle\int \dfrac{1}{x^2+49}\,dx=\dfrac{1}{7}\arctan\dfrac{x}{7}+c\).
Compare the numerator with the derivative of the denominator:
| \(f(x)\) | \(=\) | \(x^4+2\) |
| \(f'(x)\) | \(=\) | \(4x^3\) |
The top is exactly \(f'(x)\), so use \(\int\dfrac{f'}{f}\,dx=\ln|f|+c\):
| \(\int \dfrac{4x^3}{x^4+2}\,dx\) | \(=\) | \(\int \dfrac{f'(x)}{f(x)}\,dx\) |
| \(=\) | \(\ln|x^4+2|+c\) | |
| \(=\) | \(\ln(x^4+2)+c\) |
\(\displaystyle\int \dfrac{4x^3}{x^4+2}\,dx=\ln(x^4+2)+c\).
It is a product of \(x\) and \(\sin x\) with no inner-derivative factor, so integrate by parts with \(u=x\), \(dv=\sin x\,dx\):
| \(u\) | \(=\) | \(x, \quad dv=\sin x\,dx\) |
| \(du\) | \(=\) | \(dx, \quad v=-\cos x\) |
Apply \(\int u\,dv=uv-\int v\,du\):
| \(\int x\sin x\,dx\) | \(=\) | \(-x\cos x-\int(-\cos x)\,dx\) |
| \(=\) | \(-x\cos x+\int \cos x\,dx\) | |
| \(=\) | \(-x\cos x+\sin x+c\) |
\(\displaystyle\int x\sin x\,dx=\sin x-x\cos x+c\).
The denominator factorises into two distinct linear factors, so split into partial fractions:
| \(x^2-4\) | \(=\) | \((x-2)(x+2)\) |
| \(\dfrac{1}{(x-2)(x+2)}\) | \(=\) | \(\dfrac{A}{x-2}+\dfrac{B}{x+2}\) |
| \(1\) | \(=\) | \(A(x+2)+B(x-2)\) |
Substitute the roots to find \(A\) and \(B\):
| \(x=2:\quad 1\) | \(=\) | \(4A \;\Rightarrow\; A=\dfrac{1}{4}\) |
| \(x=-2:\quad 1\) | \(=\) | \(-4B \;\Rightarrow\; B=-\dfrac{1}{4}\) |
Integrate each term as a logarithm:
| \(\int \dfrac{1}{x^2-4}\,dx\) | \(=\) | \(\dfrac{1}{4}\ln|x-2|-\dfrac{1}{4}\ln|x+2|+c\) |
| \(=\) | \(\dfrac{1}{4}\ln\left|\dfrac{x-2}{x+2}\right|+c\) |
\(\displaystyle\int \dfrac{1}{x^2-4}\,dx=\dfrac{1}{4}\ln\left|\dfrac{x-2}{x+2}\right|+c\).
Common pitfalls
Frequently asked questions
How do you decide which integration technique to use?
Read the integrand's shape. An inverse-trig denominator (\(\sqrt{a^2-x^2}\) or \(a^2+x^2\)) or \(\sec^2 x\) is a standard form; a numerator equal to the derivative of the denominator is the \(\ln|f|\) form; a composite times its inner derivative is substitution; a product of unlike functions is by parts; a rational function with a factorable denominator is partial fractions.
When do you use partial fractions instead of substitution?
Use partial fractions when the integrand is a rational function whose denominator factorises into distinct linear factors and the numerator is not the derivative of the denominator. If the numerator is that derivative, the quicker route is the \(\dfrac{f'(x)}{f(x)}=\ln|f(x)|\) form.
How do you know an integral needs integration by parts?
Look for a product of two unlike functions where neither is the derivative of the other, such as \(x\sin x\), \(x e^{x}\) or \(\ln x\). With no inner-derivative factor for substitution, integrate by parts using \(\int u\,dv=uv-\int v\,du\).
What does it mean to combine integration techniques?
Some integrands are not a single form, so you rewrite first and then integrate each piece. For example \(\dfrac{2x+3}{x^2+9}\) is split into an \(f'/f\) logarithm plus an arctan form, and even powers of \(\sin\) or \(\cos\) are simplified with an identity before integrating.
How do you tell an arcsin integral from an arctan integral?
A denominator of \(\sqrt{a^2-x^2}\) (a difference under a square root) integrates to \(\arcsin\dfrac{x}{a}\); a denominator of \(a^2+x^2\) (a sum, no square root) integrates to \(\dfrac{1}{a}\arctan\dfrac{x}{a}\).
Do these techniques change for a definite integral?
The method of choosing is the same. Find the anti-derivative, then substitute the limits. If you use substitution you may either change the limits to the new variable or back-substitute first; for an integral of \(|f(x)|\), split at the zeros before evaluating.