Mixed Integration Techniques
You now have a toolbox: substitution, trig identities, inverse trig forms, completing the square, log rule, and partial fractions. The challenge in this lesson is knowing which tool to reach for. We look at six integral types, develop a decision framework, and practise reading the integral to select the right technique before writing a single symbol.
Look at $\displaystyle\int \frac{x}{\sqrt{1-x^4}}\,dx$. Without working it out — which technique would you reach for first? Write your reasoning and the substitution (or identity) you'd try.
Mixed integration is a two-step skill: identify the structure, then apply the right technique. The table below is your first-read checklist.
Quick-read checklist (in order):
- Is the numerator the derivative of the denominator? → Log rule
- Is there a composite function and its derivative? → Substitution
- Does the denominator factor into $a^2\pm x^2$? → Inverse trig
- Is there a trig power ($\sin^2$, $\cos^2$, $\tan^2$)? → Trig identity
- Is the denominator an irreducible quadratic with a linear term in $x$? → Complete the square
- Rational function with factorable denominator? → Partial fractions
Key facts
- All six Integration techniques in the Module 8 toolkit: substitution, trig identities, inverse trig, CTS, log rule, partial fractions
- The six-step decision checklist for selecting the right technique
- When an integral needs two techniques applied in sequence
Concepts
- Why reading the whole integral before acting is essential
- How a substitution can transform one problem type into another (e.g. sub then inverse trig)
- How to verify answers by differentiating
Skills
- Classify an integral by type from cold (no hints)
- Carry out each technique correctly in unseen mixed problems
- Differentiate an antiderivative to verify correctness under exam conditions
Below are the six integral types with their signature patterns and results:
Type 1 — Log rule
Example: $\int \frac{2x}{x^2+1}\,dx = \ln(x^2+1)+C$.
Type 2 — $u$-substitution
Example: $\int 2x\sqrt{x^2+3}\,dx$. Let $u=x^2+3$, $du=2x\,dx$: $= \int \sqrt{u}\,du = \tfrac{2}{3}u^{3/2}+C$.
Type 3 — Inverse trig direct
Type 4 — Trig identity then integrate
Six standard types: power, $e^x$, $\sin/\cos$, $\sec^2$, $\frac{1}{\sqrt{a^2-x^2}}\to\sin^{-1}(x/a)$, $\frac{1}{a^2+x^2}\to\frac{1}{a}\tan^{-1}(x/a)$.
Pause — copy all six integral types from the quick-reference in a two-column table: pattern on the left, result on the right into your book.
Quick check: Which technique should be used first to evaluate $\displaystyle\int \frac{3x^2}{x^3+5}\,dx$?
We just saw the six-type quick-reference: power rule, exponential, $\sin/\cos$, $\sec^2/\tan$, $\sin^{-1}$ standard form, $\tan^{-1}$ standard form — each with its exact pattern and result. That raises a question: how do you handle a harder integral that doesn't immediately match any standard form but can be reduced to one by a substitution? This card answers it → make a substitution $u=g(x)$ to simplify, then recognise the transformed integrand as a standard form.
Many harder integrals require a substitution that transforms the integrand into a standard inverse trig form. The key is to spot that after the substitution, you will have $\int \frac{du}{\sqrt{a^2-u^2}}$ or $\int \frac{du}{a^2+u^2}$.
Example: $\displaystyle\int \frac{e^x}{\sqrt{1-e^{2x}}}\,dx$.
Let $u = e^x$, $du = e^x\,dx$. Then $e^{2x} = u^2$ and:
Pattern to memorise: If you see $\sqrt{1-f(x)^2}$ or $1+f(x)^2$ in the denominator and $f'(x)$ in the numerator, set $u = f(x)$.
Many harder integrals require a substitution that transforms the integrand into a standard inverse trig form. The key is to spot that after the substitution, you will have $\int \frac{du}{\sqrt{a^2-u^2}}$ or $\int \frac{du}{a^2+u^2}$.
Pause — copy the two-step approach: (1) substitute $u=g(x)$ to simplify the integrand; (2) recognise the resulting form as $\frac{1}{\sqrt{a^2-u^2}}$ or $\frac{1}{a^2+u^2}$ and apply the standard result into your book.
Did you get this? True or false: The substitution $u = \sin x$ transforms $\displaystyle\int \frac{\cos x}{1+\sin^2 x}\,dx$ into $\displaystyle\int \frac{du}{1+u^2}$, giving $\tan^{-1}(\sin x)+C$.
Worked examples · 3 in a row, reveal as you go
Find $\displaystyle\int \frac{x}{x^2+9}\,dx$.
Find $\displaystyle\int \frac{\sin x}{\sqrt{4-\cos^2 x}}\,dx$.
Evaluate $\displaystyle\int_0^2 \frac{dx}{x^2-2x+5}$.
Fill the gap: $\displaystyle\int \frac{\cos x}{1+\sin^2 x}\,dx = \tan^{-1}($ $)+C$ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\displaystyle\int \frac{2x+3}{x^2+1}\,dx = \ln(x^2+1) + 3\tan^{-1}(x)+C$.
Activities · practice with the ideas
Identify the integration technique for each integral (do not evaluate): (a) $\int \frac{6x^2}{x^3-4}\,dx$; (b) $\int \frac{1}{4+x^2}\,dx$; (c) $\int \sin^2 3x\,dx$; (d) $\int \frac{1}{x^2+2x+10}\,dx$.
Find $\displaystyle\int \frac{4x}{x^2+7}\,dx$.
Find $\displaystyle\int \frac{e^{2x}}{1+e^{4x}}\,dx$. (Hint: let $u=e^{2x}$.)
Find $\displaystyle\int \frac{x+5}{x^2+4x+13}\,dx$. (Hint: split the numerator so one part is the derivative of the denominator.)
Verify that $\dfrac{d}{dx}\!\left[-\sin^{-1}\!\left(\dfrac{\cos x}{2}\right)\right] = \dfrac{\sin x}{\sqrt{4-\cos^2 x}}$.
Odd one out: Three of these integrals correctly matched to their technique. Which matching is WRONG?
Earlier you wrote down the technique you'd use for $\displaystyle\int \frac{x}{\sqrt{1-x^4}}\,dx$.
The cleanest path: substitute $u = x^2$, $du = 2x\,dx$, to get $\frac{1}{2}\int \frac{du}{\sqrt{1-u^2}} = \frac{1}{2}\sin^{-1}(x^2)+C$. This is a classic "hidden $\sin^{-1}$" problem — the $x^4 = (x^2)^2$ in the denominator is the clue. The lesson: always check whether a substitution can turn $f(x)^n$ terms into $(u)^n$, revealing the standard form.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find $\displaystyle\int \frac{e^{2x}}{1+e^{4x}}\,dx$, showing the substitution clearly. (2 marks)
Q2. Find $\displaystyle\int \frac{x+5}{x^2+4x+13}\,dx$. Show all steps. (3 marks)
Q3. Evaluate $\displaystyle\int_0^{\pi/2} \frac{\sin x}{4+\cos^2 x}\,dx$, justifying your technique choice. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. (a) log rule; (b) inverse tan; (c) trig identity ($\cos^2 3x = \frac{1+\cos 6x}{2}$); (d) completing the square.
2. $2\ln(x^2+7)+C$.
3. $\frac{1}{2}\tan^{-1}(e^{2x})+C$.
4. $x+5 = \frac{1}{2}(2x+4)+3$; CTS: $x^2+4x+13=(x+2)^2+9$; answer: $\frac{1}{2}\ln(x^2+4x+13)+\tan^{-1}\!\frac{x+2}{3}+C$.
5. Chain rule: $\frac{d}{dx}\!\left[-\sin^{-1}\!\frac{\cos x}{2}\right] = -\frac{1}{\sqrt{1-(\cos x/2)^2}}\cdot\frac{-\sin x}{2} = \frac{\sin x}{2\sqrt{1-\cos^2 x/4}} = \frac{\sin x}{2\cdot\frac{1}{2}\sqrt{4-\cos^2 x}} = \frac{\sin x}{\sqrt{4-\cos^2 x}}$ ✓
Q1 (2 marks): Let $u = e^{2x}$, $du = 2e^{2x}\,dx$ [1]. $\int \frac{e^{2x}\,dx}{1+e^{4x}} = \frac{1}{2}\int \frac{du}{1+u^2} = \frac{1}{2}\tan^{-1}(e^{2x})+C$ [1].
Q2 (3 marks): $x+5 = \frac{1}{2}(2x+4)+3$ [1]. $\int \frac{x+5}{x^2+4x+13}dx = \frac{1}{2}\ln(x^2+4x+13) + 3\int \frac{dx}{(x+2)^2+9}$ [1]. $= \frac{1}{2}\ln(x^2+4x+13)+\tan^{-1}\!\frac{x+2}{3}+C$ [1].
Q3 (3 marks): Technique: sub $u = \cos x$, $du = -\sin x\,dx$ [1]. Limits: $x=0 \Rightarrow u=1$; $x=\frac{\pi}{2} \Rightarrow u=0$. $\int_0^{\pi/2}\frac{\sin x\,dx}{4+\cos^2 x} = -\int_1^0\frac{du}{4+u^2} = \int_0^1\frac{du}{4+u^2}$ [1]. $= \left[\frac{1}{2}\tan^{-1}\!\frac{u}{2}\right]_0^1 = \frac{1}{2}\tan^{-1}\!\frac{1}{2}$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering mixed integration technique questions. Lighter alternative to the boss.
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