Calculus · Paper 1
Integration — Part 2
The 8 Methods · Higher Level · Tap NEXT to begin
Section 1 of 13
Method 1 — Multiply out
Part 1 ended with the integral sign. Now we need to actually compute $\displaystyle\int f(x)\,dx$ for every kind of $f$ you'll meet on the paper.
There are 8 methods. Eight ways the question can be dressed up. The trick is recognising which one you're looking at.
The 8 methods at a glance
1 · Multiply out · brackets times brackets
2 · Factors · cancel a common factor first
3 · Indices · rewrite as $x^{n}$ (negative or fractional powers)
4 · Split the fraction · one term on the bottom → split across
5 · Tables · standard integrals from the formula booklet
6 · $e^{x}$ and $a^{x}$ · exponentials
7 · Trig → Product → Sum · convert before integrating
8 · Special case · reverse product rule (for $xe^{x}$ and $\ln x$)
2 · Factors · cancel a common factor first
3 · Indices · rewrite as $x^{n}$ (negative or fractional powers)
4 · Split the fraction · one term on the bottom → split across
5 · Tables · standard integrals from the formula booklet
6 · $e^{x}$ and $a^{x}$ · exponentials
7 · Trig → Product → Sum · convert before integrating
8 · Special case · reverse product rule (for $xe^{x}$ and $\ln x$)
Method 1 — Multiply out. If you see two brackets multiplied together, expand them first. Never try to integrate the brackets directly.
(i) $\displaystyle\int (x+3)(x+5)\,dx$
Multiply out: $(x+3)(x+5) = x^{2} + 8x + 15$
$\displaystyle\int (x^{2} + 8x + 15)\,dx$
$= \dfrac{x^{3}}{3} + \dfrac{8x^{2}}{2} + 15x + c$
$= \dfrac{x^{3}}{3} + 4x^{2} + 15x + c$
(ii) $\displaystyle\int (x+1)^{2}\,dx$ — with a wrong attempt
A trap students fall into:
$\displaystyle\int (x+1)^{2}\,dx = \dfrac{(x+1)^{3}}{3} + c$ ✗
That's not how integration works. The power rule only works on bare powers of $x$, not on brackets. Always expand first.
$(x+1)^{2} = x^{2} + 2x + 1$
$\displaystyle\int (x^{2} + 2x + 1)\,dx$
$= \dfrac{x^{3}}{3} + \dfrac{2x^{2}}{2} + x + c$
$= \dfrac{x^{3}}{3} + x^{2} + x + c$
You try
Find $\displaystyle\int (x-4)(x+2)\,dx$.
Multiply out first. Then integrate term by term.
$(x-4)(x+2) = x^{2} - 2x - 8$
$\displaystyle\int (x^{2} - 2x - 8)\,dx$
$= \dfrac{x^{3}}{3} - \dfrac{2x^{2}}{2} - 8x + c$
$= \dfrac{x^{3}}{3} - x^{2} - 8x + c$
$\dfrac{x^{3}}{3} - x^{2} - 8x + c$
Section 2 of 13
Method 2 — Factors
If you see a fraction with a polynomial top and a polynomial bottom, look for a common factor that cancels. Once you cancel, the rest is standard.
(i) $\displaystyle\int \dfrac{x^{3} - 27}{x - 3}\,dx$
The top is a difference of cubes: $a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2})$.
$x^{3} - 27 = x^{3} - 3^{3} = (x-3)(x^{2} + 3x + 9)$
$\displaystyle\int \dfrac{(x-3)(x^{2} + 3x + 9)}{x - 3}\,dx$
Cancel $(x-3)$ top and bottom:
$\displaystyle\int (x^{2} + 3x + 9)\,dx$
$= \dfrac{x^{3}}{3} + \dfrac{3x^{2}}{2} + 9x + c$
(ii) $\displaystyle\int \dfrac{x^{2} - 9}{x - 3}\,dx$
Easier case — difference of squares.
$x^{2} - 9 = (x-3)(x+3)$
$\displaystyle\int \dfrac{(x-3)(x+3)}{x - 3}\,dx = \displaystyle\int (x+3)\,dx$
$= \dfrac{x^{2}}{2} + 3x + c$
The factoring cheatsheet
$a^{2} - b^{2} = (a-b)(a+b)$ · difference of squares
$a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2})$ · difference of cubes
$a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$ · sum of cubes
$a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2})$ · difference of cubes
$a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$ · sum of cubes
You try
Find $\displaystyle\int \dfrac{x^{3} - 8}{x - 2}\,dx$.
Difference of cubes: $x^{3} - 2^{3}$. Factor it then cancel.
$x^{3} - 8 = (x-2)(x^{2} + 2x + 4)$
$\displaystyle\int \dfrac{(x-2)(x^{2} + 2x + 4)}{x - 2}\,dx = \displaystyle\int (x^{2} + 2x + 4)\,dx$
$= \dfrac{x^{3}}{3} + \dfrac{2x^{2}}{2} + 4x + c$
$= \dfrac{x^{3}}{3} + x^{2} + 4x + c$
$\dfrac{x^{3}}{3} + x^{2} + 4x + c$
Section 3 of 13
Method 3 — Indices
Anything that isn't already in $x^{n}$ form — rewrite it. Then the power rule does the work.
Must learn — the rewriting
$\dfrac{1}{x^{n}} = x^{-n}$ · flip $\to$ negative power
$\sqrt{x} = x^{1/2}$ · root $\to$ fractional power
$\sqrt[n]{x} = x^{1/n}$ · nth root $\to$ $1/n$ power
$\sqrt{x} = x^{1/2}$ · root $\to$ fractional power
$\sqrt[n]{x} = x^{1/n}$ · nth root $\to$ $1/n$ power
(i) $\displaystyle\int \dfrac{1}{x^{2}}\,dx$
$\displaystyle\int \dfrac{1}{x^{2}}\,dx = \displaystyle\int x^{-2}\,dx$
Increase power: $-2 + 1 = -1$
$= \dfrac{x^{-1}}{-1} + c$
$= -\dfrac{1}{x} + c$
(ii) $\displaystyle\int \sqrt{x}\,dx$
$\displaystyle\int \sqrt{x}\,dx = \displaystyle\int x^{1/2}\,dx$
Increase power: $\dfrac{1}{2} + 1 = \dfrac{3}{2}$
$= \dfrac{x^{3/2}}{\tfrac{3}{2}} + c$
Dividing by $\tfrac{3}{2}$ = multiplying by $\tfrac{2}{3}$:
$= \dfrac{2}{3}\, x^{3/2} + c$
$= \dfrac{2}{3}\, x\sqrt{x} + c$ (since $x^{3/2} = x \cdot x^{1/2} = x\sqrt{x}$)
(iii) $\displaystyle\int \dfrac{dx}{\sqrt{x}}$
$\displaystyle\int \dfrac{1}{\sqrt{x}}\,dx = \displaystyle\int x^{-1/2}\,dx$
Increase power: $-\dfrac{1}{2} + 1 = \dfrac{1}{2}$
$= \dfrac{x^{1/2}}{\tfrac{1}{2}} + c = 2x^{1/2} + c$
$= 2\sqrt{x} + c$
You try
Find $\displaystyle\int \dfrac{1}{x^{3}}\,dx$.
Rewrite as $x^{-3}$. New power: $-3 + 1 = -2$.
$\displaystyle\int x^{-3}\,dx = \dfrac{x^{-2}}{-2} + c$
$= -\dfrac{1}{2x^{2}} + c$
$-\dfrac{1}{2x^{2}} + c$
You try — one more
Find $\displaystyle\int \sqrt[3]{x}\,dx$ (cube root of $x$).
$\sqrt[3]{x} = x^{1/3}$. Increase: $\tfrac{1}{3} + 1 = \tfrac{4}{3}$.
$\displaystyle\int x^{1/3}\,dx = \dfrac{x^{4/3}}{\tfrac{4}{3}} + c$
$= \dfrac{3}{4}\, x^{4/3} + c$
$\dfrac{3}{4}\, x^{4/3} + c$
Section 4 of 13
Method 4 — Split the fraction
One thing on the bottom, many on the top. Split the fraction across the top — every term gets the same denominator.
When it's allowed
$\dfrac{a + b + c}{d} = \dfrac{a}{d} + \dfrac{b}{d} + \dfrac{c}{d}$ ✓
Only when the single thing is on the bottom. Never the other way around — splitting across the bottom is a famous mistake.
Only when the single thing is on the bottom. Never the other way around — splitting across the bottom is a famous mistake.
(i) $\displaystyle\int \dfrac{x^{2} + 3x + 1}{x}\,dx$
$\displaystyle\int \left(\dfrac{x^{2}}{x} + \dfrac{3x}{x} + \dfrac{1}{x}\right) dx$
$= \displaystyle\int \left(x + 3 + \dfrac{1}{x}\right) dx$
$= \dfrac{x^{2}}{2} + 3x + \ln x + c$ ($\ln x$ comes from the tables — next section)
Watch the trap. The last term $\dfrac{1}{x}$ does not integrate by the power rule. There's a special rule for it — see Section 6.
You try
Find $\displaystyle\int \dfrac{2x^{2} + 4x + 6}{2x}\,dx$.
Split across the top. $\dfrac{2x^{2}}{2x} = x$, $\dfrac{4x}{2x} = 2$, $\dfrac{6}{2x} = \dfrac{3}{x}$.
$\displaystyle\int \left(\dfrac{2x^{2}}{2x} + \dfrac{4x}{2x} + \dfrac{6}{2x}\right) dx$
$= \displaystyle\int \left(x + 2 + \dfrac{3}{x}\right) dx$
$= \dfrac{x^{2}}{2} + 2x + 3\ln x + c$
$\dfrac{x^{2}}{2} + 2x + 3\ln x + c$
Section 5 of 13
Method 5 — From the Tables
Some integrals aren't done by rule — they're looked up in the formula booklet. The relevant page is the integration table. Worth knowing the most common ones by heart anyway.
| $f(x)$ | $\displaystyle\int f(x)\,dx$ |
|---|---|
| $x^{n}$ ($n$ ≠ $-1$) | $\dfrac{x^{n+1}}{n+1}$ |
| $\dfrac{1}{x}$ | $\ln |x|$ |
| $e^{x}$ | $e^{x}$ |
| $e^{ax}$ | $\dfrac{1}{a}\,e^{ax}$ |
| $a^{x}$ | $\dfrac{a^{x}}{\ln a}$ |
| $\cos x$ | $\sin x$ |
| $\sin x$ | $-\cos x$ |
| $\tan x$ | $\ln |\sec x|$ |
All worth knowing. The trig ones and the inverse trig ones we'll handle in their own sections — first let's nail $1/x$ and the exponentials.
Section 6 of 13
The $\dfrac{1}{x}$ rule — $\ln x$
This one's special. The power rule fails on $\dfrac{1}{x}$ because:
$\displaystyle\int \dfrac{1}{x}\,dx = \displaystyle\int x^{-1}\,dx \stackrel{?}{=} \dfrac{x^{0}}{0}$ ✗ (dividing by 0 — undefined)
So the power rule has a hole exactly when $n = -1$. The Tables fill that hole:
Must learn
$\displaystyle\int \dfrac{1}{x}\,dx = \ln x + c$
The exam tables list $\ln |x|$. For positive $x$ — the only kind that comes up at HL — it's just $\ln x$.
(i) $\displaystyle\int \dfrac{x^{2} + 3x + 5}{x}\,dx$ — combined with Split
Split first: $\displaystyle\int \left(\dfrac{x^{2}}{x} + \dfrac{3x}{x} + \dfrac{5}{x}\right) dx$
$= \displaystyle\int \left(x + 3 + \dfrac{5}{x}\right) dx$ (note $\dfrac{5}{x} = 5 \cdot \dfrac{1}{x}$, so the $5$ travels out)
$= \dfrac{x^{2}}{2} + 3x + 5\ln x + c$
(ii) $\displaystyle\int \dfrac{7}{x}\,dx$
The $7$ is a constant — pulls out front:
$= 7 \displaystyle\int \dfrac{1}{x}\,dx = 7\ln x + c$
You try
Find $\displaystyle\int \dfrac{x + 2}{x}\,dx$.
Split: $\dfrac{x}{x} + \dfrac{2}{x} = 1 + \dfrac{2}{x}$.
$\displaystyle\int \left(\dfrac{x}{x} + \dfrac{2}{x}\right) dx = \displaystyle\int \left(1 + \dfrac{2}{x}\right) dx$
$= x + 2\ln x + c$
$x + 2\ln x + c$
Section 7 of 13
Tables — inverse trig forms
Two more from the tables, and these have to come from the tables — you'd never spot them otherwise.
Must learn — straight from the tables
$\displaystyle\int \dfrac{1}{x^{2} + a^{2}}\,dx = \dfrac{1}{a}\,\tan^{-1}\!\dfrac{x}{a} + c$
$\displaystyle\int \dfrac{1}{\sqrt{a^{2} - x^{2}}}\,dx = \sin^{-1}\!\dfrac{x}{a} + c$
$\displaystyle\int \dfrac{1}{\sqrt{a^{2} - x^{2}}}\,dx = \sin^{-1}\!\dfrac{x}{a} + c$
$a \in \mathbb{R}$ — any positive constant.
(i) $\displaystyle\int \dfrac{1}{x^{2} + 16}\,dx$
A common trap: try to use indices.
$\displaystyle\int (x^{2} + 16)^{-1} dx \stackrel{?}{=} \dfrac{(x^{2} + 16)^{0}}{0}$ ✗ (power rule doesn't apply to brackets)
Another trap: try to factorise.
$x^{2} + 16$ ≠ $(x-4)(x+4)$ ✗ (that would give $x^{2} - 16$, sum is irreducible over $\mathbb{R}$)
So go to the tables. Match the form $\dfrac{1}{x^{2} + a^{2}}$ — here $a^{2} = 16$, so $a = 4$.
$\displaystyle\int \dfrac{1}{x^{2} + 16}\,dx = \displaystyle\int \dfrac{1}{x^{2} + 4^{2}}\,dx$
$= \dfrac{1}{4}\tan^{-1}\!\dfrac{x}{4} + c$
(ii) $\displaystyle\int \dfrac{1}{\sqrt{25 - x^{2}}}\,dx$
Match $\dfrac{1}{\sqrt{a^{2} - x^{2}}}$ — here $a^{2} = 25$, so $a = 5$.
$\displaystyle\int \dfrac{1}{\sqrt{5^{2} - x^{2}}}\,dx = \sin^{-1}\!\dfrac{x}{5} + c$
You try
Find $\displaystyle\int \dfrac{1}{x^{2} + 9}\,dx$.
$a^{2} = 9$, so $a = 3$.
$\displaystyle\int \dfrac{1}{x^{2} + 3^{2}}\,dx$
$= \dfrac{1}{3}\tan^{-1}\!\dfrac{x}{3} + c$
$\dfrac{1}{3}\tan^{-1}\!\dfrac{x}{3} + c$
You try — one more
Find $\displaystyle\int \dfrac{1}{\sqrt{49 - x^{2}}}\,dx$.
$a^{2} = 49$, so $a = 7$. Use $\sin^{-1}\!\dfrac{x}{a}$.
$\displaystyle\int \dfrac{1}{\sqrt{7^{2} - x^{2}}}\,dx$
$= \sin^{-1}\!\dfrac{x}{7} + c$
$\sin^{-1}\!\dfrac{x}{7} + c$
Section 8 of 13
Method 6 — $e^{x}$, $e^{ax}$ and $a^{x}$
From the tables:
Must learn
$\displaystyle\int e^{x}\,dx = e^{x} + c$
$\displaystyle\int e^{ax}\,dx = \dfrac{1}{a}\,e^{ax} + c$
$\displaystyle\int a^{x}\,dx = \dfrac{a^{x}}{\ln a} + c$
$\displaystyle\int e^{ax}\,dx = \dfrac{1}{a}\,e^{ax} + c$
$\displaystyle\int a^{x}\,dx = \dfrac{a^{x}}{\ln a} + c$
(i) $\displaystyle\int e^{5x}\,dx$
$a = 5$, so:
$= \dfrac{1}{5}\,e^{5x} + c$
(ii) $\displaystyle\int \dfrac{1}{e^{3x}}\,dx$ — flip first
$\dfrac{1}{e^{3x}} = e^{-3x}$ (flip $\to$ negative power, like indices)
$\displaystyle\int e^{-3x}\,dx = -\dfrac{1}{3}\,e^{-3x} + c$ ($a = -3$)
$= -\dfrac{1}{3\,e^{3x}} + c$ (write it back as a fraction if you like)
(iii) $\displaystyle\int 10^{x}\,dx$
$a = 10$:
$= \dfrac{10^{x}}{\ln 10} + c$
You try
Find $\displaystyle\int e^{2x}\,dx$.
$a = 2$, so divide by $2$.
$\dfrac{1}{2}\,e^{2x} + c$
$\dfrac{1}{2}\,e^{2x} + c$
You try — one more
Find $\displaystyle\int \dfrac{1}{e^{4x}}\,dx$.
Flip first: $\dfrac{1}{e^{4x}} = e^{-4x}$. Then $a = -4$.
$\displaystyle\int e^{-4x}\,dx = -\dfrac{1}{4}\,e^{-4x} + c$
$= -\dfrac{1}{4\,e^{4x}} + c$
$-\dfrac{1}{4\,e^{4x}} + c$
Section 9 of 13
Method 7a — Trig with a coefficient
From the tables, $\displaystyle\int \sin x\,dx = -\cos x + c$ and $\displaystyle\int \cos x\,dx = \sin x + c$. But what if the angle is $3x$ or $5x$ instead of just $x$?
Must learn — the $\dfrac{1}{a}$ rule
$\displaystyle\int \cos ax\,dx = \dfrac{1}{a}\sin ax + c$
$\displaystyle\int \sin ax\,dx = -\dfrac{1}{a}\cos ax + c$
$\displaystyle\int \sin ax\,dx = -\dfrac{1}{a}\cos ax + c$
Integrate as normal, then divide by the coefficient $a$.
(i) $\displaystyle\int \cos 3x\,dx$
$= \dfrac{1}{3}\sin 3x + c$
(ii) $\displaystyle\int \sin 7x\,dx$
$= -\dfrac{1}{7}\cos 7x + c$
(iii) Sums — $\displaystyle\int (\sin 5x - \cos 3x)\,dx$
Handle each term separately.
$\displaystyle\int \sin 5x\,dx = -\dfrac{1}{5}\cos 5x$
$\displaystyle\int \cos 3x\,dx = \dfrac{1}{3}\sin 3x$
$\displaystyle\int (\sin 5x - \cos 3x)\,dx = -\dfrac{1}{5}\cos 5x - \dfrac{1}{3}\sin 3x + c$
You try
Find $\displaystyle\int (\sin 7x - \cos 9x)\,dx$.
Term by term. $\sin 7x \to -\dfrac{1}{7}\cos 7x$. $\cos 9x \to \dfrac{1}{9}\sin 9x$.
$-\dfrac{1}{7}\cos 7x - \dfrac{1}{9}\sin 9x + c$
$-\dfrac{1}{7}\cos 7x - \dfrac{1}{9}\sin 9x + c$
Section 10 of 13
Method 7b — Trig: Products $\to$ Sums
You cannot integrate a product of trig functions directly. $\displaystyle\int \cos 7x \cos x\,dx$ is not $\displaystyle\int \cos 7x \,dx \times \displaystyle\int \cos x\,dx$. There's no product rule for integration.
The trick: convert the product into a sum first, using the formulas from the tables. Then integrate term by term.
Products $\to$ Sums · from the tables
$2\cos A \cos B = \cos(A-B) + \cos(A+B)$
$2\sin A \cos B = \sin(A+B) + \sin(A-B)$
$2\sin A \sin B = \cos(A-B) - \cos(A+B)$
$2\cos A \sin B = \sin(A+B) - \sin(A-B)$
$2\sin A \cos B = \sin(A+B) + \sin(A-B)$
$2\sin A \sin B = \cos(A-B) - \cos(A+B)$
$2\cos A \sin B = \sin(A+B) - \sin(A-B)$
Each one starts with 2. So if your integrand isn't $2\times\!$something, multiply by $\dfrac{2}{2}$ — keep the $\dfrac{1}{2}$ out front, the $2$ on the inside.
(i) $\displaystyle\int 2\sin 3x \sin x\,dx$ — matches the formula exactly
Use $2\sin A \sin B = \cos(A-B) - \cos(A+B)$ with $A = 3x$, $B = x$:
$2\sin 3x \sin x = \cos(3x - x) - \cos(3x + x)$
$= \cos 2x - \cos 4x$
$\displaystyle\int (\cos 2x - \cos 4x)\,dx$
$= \dfrac{1}{2}\sin 2x - \dfrac{1}{4}\sin 4x + c$
(ii) $\displaystyle\int \cos 7x \cos x\,dx$ — need to multiply by $\dfrac{2}{2}$ first
No $2$ in front, so put one there. Compensate with $\dfrac{1}{2}$ outside the integral.
$\displaystyle\int \cos 7x \cos x\,dx = \dfrac{1}{2}\displaystyle\int 2\cos 7x \cos x\,dx$
Now apply $2\cos A \cos B = \cos(A-B) + \cos(A+B)$ with $A = 7x$, $B = x$:
$= \dfrac{1}{2}\displaystyle\int \big[\cos 6x + \cos 8x\big]\,dx$
$= \dfrac{1}{2}\left[\dfrac{1}{6}\sin 6x + \dfrac{1}{8}\sin 8x\right] + c$
$= \dfrac{1}{12}\sin 6x + \dfrac{1}{16}\sin 8x + c$
(iii) $\displaystyle\int 3\sin 5x \cos x\,dx$ — numerical coefficient out front
Pull the $3$ out, then handle the product with $2\sin A \cos B = \sin(A+B) + \sin(A-B)$. Multiply by $\dfrac{2}{2}$ to get the leading $2$:
$3\displaystyle\int \sin 5x \cos x\,dx = \dfrac{3}{2}\displaystyle\int 2\sin 5x \cos x\,dx$
$= \dfrac{3}{2}\displaystyle\int \big[\sin 6x + \sin 4x\big]\,dx$
$= \dfrac{3}{2}\left[-\dfrac{1}{6}\cos 6x - \dfrac{1}{4}\cos 4x\right] + c$
$= -\dfrac{1}{4}\cos 6x - \dfrac{3}{8}\cos 4x + c$
You try
Find $\displaystyle\int 2\cos 5x \cos 3x\,dx$.
Matches $2\cos A \cos B$. $A = 5x$, $B = 3x$. $A-B = 2x$, $A+B = 8x$.
$2\cos 5x \cos 3x = \cos 2x + \cos 8x$
$\displaystyle\int (\cos 2x + \cos 8x)\,dx$
$= \dfrac{1}{2}\sin 2x + \dfrac{1}{8}\sin 8x + c$
$\dfrac{1}{2}\sin 2x + \dfrac{1}{8}\sin 8x + c$
You try — one more
Find $\displaystyle\int \sin 11x \cos 3x\,dx$.
No $2$ in front — multiply by $\dfrac{2}{2}$. Then $2\sin A \cos B$ with $A=11x$, $B=3x$.
$\dfrac{1}{2}\displaystyle\int 2\sin 11x \cos 3x\,dx$
$= \dfrac{1}{2}\displaystyle\int \big[\sin 14x + \sin 8x\big]\,dx$
$= \dfrac{1}{2}\left[-\dfrac{1}{14}\cos 14x - \dfrac{1}{8}\cos 8x\right] + c$
$= -\dfrac{1}{28}\cos 14x - \dfrac{1}{16}\cos 8x + c$
$-\dfrac{1}{28}\cos 14x - \dfrac{1}{16}\cos 8x + c$
Section 11 of 13
Method 8 — Special Case (reverse product rule)
Two integrals you'll meet that none of the other methods touch: $\displaystyle\int xe^{x}\,dx$ and $\displaystyle\int \ln x\,dx$.
There's a clever trick. The exam gives you a helper part — differentiate something first, then rearrange that result to make the integral fall out.
The recipe
(1) Differentiate the function the question hands you (usually $y = xe^{x}$ or $f(x) = x\ln x$) using the product rule.
(2) Spot the awkward integrand somewhere inside that derivative.
(3) Rearrange to isolate it, then integrate both sides.
(2) Spot the awkward integrand somewhere inside that derivative.
(3) Rearrange to isolate it, then integrate both sides.
(i) $\displaystyle\int xe^{x}\,dx$ — given $y = xe^{x}$
Part 1 — differentiate. Product rule on $y = xe^{x}$ (first $\times$ derivative of second + second $\times$ derivative of first):
$\dfrac{dy}{dx} = x \cdot e^{x} + e^{x} \cdot 1 = xe^{x} + e^{x}$
Part 2 — spot it. Look — there's $xe^{x}$ sitting right inside the derivative. Rearrange:
$xe^{x} = \dfrac{dy}{dx} - e^{x}$
Part 3 — integrate both sides.
$\displaystyle\int xe^{x}\,dx = \displaystyle\int \left(\dfrac{dy}{dx} - e^{x}\right) dx$
$= y - e^{x} + c$ (integrating $\dfrac{dy}{dx}$ gives back $y$)
$= xe^{x} - e^{x} + c$ (sub $y = xe^{x}$ back in)
(ii) $\displaystyle\int \ln x\,dx$ — given $f(x) = x \ln x$
Part 1 — differentiate. Product rule on $f(x) = x \ln x$:
$f'(x) = x \cdot \dfrac{1}{x} + \ln x \cdot 1 = 1 + \ln x$
Part 2 — spot it. $\ln x$ is right there. Rearrange:
$\ln x = f'(x) - 1$
Part 3 — integrate.
$\displaystyle\int \ln x\,dx = \displaystyle\int \big(f'(x) - 1\big)\,dx$
$= f(x) - x + c$
$= x\ln x - x + c$
Memorise the results
$\displaystyle\int xe^{x}\,dx = xe^{x} - e^{x} + c$
$\displaystyle\int \ln x\,dx = x\ln x - x + c$
$\displaystyle\int \ln x\,dx = x\ln x - x + c$
In the exam the helper part is given. But the answer is the same every time — worth knowing both off by heart.
You try
Given $y = x\cos x$, find $\dfrac{dy}{dx}$ and use it to find $\displaystyle\int x\sin x\,dx$.
Product rule on $x\cos x$: $\cos x \cdot 1 + x \cdot (-\sin x) = \cos x - x\sin x$. Rearrange for $x\sin x$, then integrate.
Part 1: $\dfrac{dy}{dx} = \cos x - x\sin x$
Part 2: $x\sin x = \cos x - \dfrac{dy}{dx}$
Part 3: $\displaystyle\int x\sin x\,dx = \displaystyle\int \cos x\,dx - \displaystyle\int \dfrac{dy}{dx}\,dx$
$= \sin x - y + c$
$= \sin x - x\cos x + c$
$\sin x - x\cos x + c$
Section 12 of 13
The 8 Methods — how to spot which
In the exam, the question won't say "use Method 4". You have to recognise which method applies. The clue is always in the shape of the integrand.
The checklist — if you see this, do this
1 · Multiply out — if you see brackets multiplied: $(x+a)(x+b)$ or $(x+a)^{2}$ → expand first
2 · Factors — polynomial $\div$ polynomial with a common factor → factor top, cancel
3 · Indices — $\dfrac{1}{x^{n}}$, $\sqrt{x}$, roots, or fractions involving powers → rewrite as $x^{n}$
4 · Split the fraction — sum on top, single thing on bottom → split across
5 · Tables — $\dfrac{1}{x^{2}+a^{2}}$ or $\dfrac{1}{\sqrt{a^{2}-x^{2}}}$ → $\tan^{-1}\!\dfrac{x}{a}$ or $\sin^{-1}\!\dfrac{x}{a}$
And the standard ones: $\dfrac{1}{x} \to \ln x$, $\sin x \to -\cos x$, etc.
6 · $e^{x}$ and $a^{x}$ — an exponential of any base → the table rules with $\dfrac{1}{a}$ if it's $e^{ax}$
7 · Trig with coefficient or product — $\cos ax$, $\sin ax$ → $\dfrac{1}{a}$ rule. Two trig functions multiplied → products-to-sums first
8 · Special case — $xe^{x}$ or $\ln x$ (and similar) → the exam gives a "find $\dfrac{dy}{dx}$" first part, use it
2 · Factors — polynomial $\div$ polynomial with a common factor → factor top, cancel
3 · Indices — $\dfrac{1}{x^{n}}$, $\sqrt{x}$, roots, or fractions involving powers → rewrite as $x^{n}$
4 · Split the fraction — sum on top, single thing on bottom → split across
5 · Tables — $\dfrac{1}{x^{2}+a^{2}}$ or $\dfrac{1}{\sqrt{a^{2}-x^{2}}}$ → $\tan^{-1}\!\dfrac{x}{a}$ or $\sin^{-1}\!\dfrac{x}{a}$
And the standard ones: $\dfrac{1}{x} \to \ln x$, $\sin x \to -\cos x$, etc.
6 · $e^{x}$ and $a^{x}$ — an exponential of any base → the table rules with $\dfrac{1}{a}$ if it's $e^{ax}$
7 · Trig with coefficient or product — $\cos ax$, $\sin ax$ → $\dfrac{1}{a}$ rule. Two trig functions multiplied → products-to-sums first
8 · Special case — $xe^{x}$ or $\ln x$ (and similar) → the exam gives a "find $\dfrac{dy}{dx}$" first part, use it
The pre-exam ritual. Before integrating anything, read the question and ask: which of the 8 am I looking at? Once you've named it, you know what to do.
Some questions combine two methods. The classic is Split + the $1/x$ rule, like $\displaystyle\int \dfrac{x^{2} + 5}{x}\,dx$. Split first (Method 4), then $\dfrac{5}{x}$ needs $\ln x$ from the Tables (Method 5).
You try — recognition test
For each of the following, name the method (1 – 8) you would use. You don't need to compute — just identify.
(a) $\displaystyle\int (x-1)(x+4)\,dx$
(b) $\displaystyle\int \dfrac{1}{x^{2}+25}\,dx$
(c) $\displaystyle\int \dfrac{x^{4} + 2x}{x}\,dx$
(d) $\displaystyle\int e^{7x}\,dx$
(e) $\displaystyle\int 2\sin 5x \sin 3x\,dx$
(a) $\displaystyle\int (x-1)(x+4)\,dx$
(b) $\displaystyle\int \dfrac{1}{x^{2}+25}\,dx$
(c) $\displaystyle\int \dfrac{x^{4} + 2x}{x}\,dx$
(d) $\displaystyle\int e^{7x}\,dx$
(e) $\displaystyle\int 2\sin 5x \sin 3x\,dx$
Look at the shape of each integrand. What's the giveaway in each?
(a) Brackets multiplied → Method 1 (Multiply out)
(b) Matches $\dfrac{1}{x^{2}+a^{2}}$ → Method 5 (Tables), with $a=5$
(c) One thing on the bottom, sum on top → Method 4 (Split)
(d) Exponential $e^{ax}$ → Method 6
(e) Two trig functions multiplied → Method 7 (Products $\to$ Sums)
(a) 1 · (b) 5 · (c) 4 · (d) 6 · (e) 7
Section 13 of 13
What's next
You now have the full toolkit for indefinite integration. Whatever the exam throws at you under the integral sign, one of the 8 methods will crack it.
Coming next in Integration — Part 3: we put limits on the integral. The $+c$ disappears, and the answer becomes a number — the area under the curve. That's where the long question lives. Every Paper 1 in recent years has had an integration area question worth serious marks. Get this fluent and that question is yours.
That's the 8 Methods.
Multiply out, Factors, Indices, Split, Tables, $e^{x}$, Trig (Products $\to$ Sums), Special Case. Recognise the shape, pick the method, compute. Next: definite integrals and the area under the curve.