Integration — The Table and the Four Methods

Integration is anti-differentiation: $\int f(x)\,dx = F(x) + C$ where $F' = f$, and the constant $C$ is a mark on the board scheme — omit it and pay.

The standard table (both directions of the derivative table): $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ ($n \neq -1$); $\int \frac{1}{x}dx = \log\vert x \vert + C$; $\int e^x dx = e^x + C$; $\int a^x dx = \frac{a^x}{\log a} + C$; $\int \sin x\,dx = -\cos x + C$; $\int \cos x\,dx = \sin x + C$; $\int \sec^2 x\,dx = \tan x + C$; $\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C$; $\int \frac{dx}{1+x^2} = \tan^{-1}x + C$. And the derived trio: $\int\tan x\,dx = \log\vert\sec x\vert + C$, $\int\cot x\,dx = \log\vert\sin x\vert + C$, $\int\sec x\,dx = \log\vert\sec x + \tan x\vert + C$.

Method 1 — Substitution: when the integrand contains a function and its derivative, set $t$ equal to the inner function. $\int \frac{2x}{1+x^2}dx$: put $t = 1 + x^2$, $dt = 2x\,dx$, giving $\log(1 + x^2) + C$. The shape $\int \frac{f'(x)}{f(x)}dx = \log\vert f(x)\vert + C$ is worth recognising on sight.

Method 2 — Special quadratic forms: $$\int\frac{dx}{x^2 - a^2} = \frac{1}{2a}\log\left\vert\frac{x-a}{x+a}\right\vert + C, \quad \int\frac{dx}{a^2 - x^2} = \frac{1}{2a}\log\left\vert\frac{a+x}{a-x}\right\vert + C, \quad \int\frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$$ $$\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C, \qquad \int\frac{dx}{\sqrt{x^2 \pm a^2}} = \log\left\vert x + \sqrt{x^2 \pm a^2}\right\vert + C$$ For a general quadratic, complete the square first: $x^2 + 2x + 2 = (x+1)^2 + 1$, then apply the matching form.

Method 3 — Integration by parts: $\int u\,v\,dx = u\int v\,dx - \int\left(u'\int v\,dx\right)dx$. Choose $u$ by ILATE (Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential) — the earlier in the list, the better a $u$. Two must-know products: $\int\log x\,dx = x\log x - x + C$ (take $v = 1$), and the shortcut $$\int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) + C$$

Method 4 — Partial fractions (proper rational functions, distinct/repeated linear and quadratic factors): split, then integrate term by term — $\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2}$ is the template.

Definite Integrals and Their Properties

Fundamental theorem: $\int_a^b f(x)\,dx = F(b) - F(a)$ — no constant $C$ in definite answers. Under substitution, change the limits along with the variable.

The property toolkit:

1. $\int_a^b f(x)\,dx = \int_a^b f(t)\,dt$ (dummy variable) and $\int_a^b = -\int_b^a$.
2. $\int_a^b = \int_a^c + \int_c^b$ (splitting — essential for modulus integrands).
3. King's property: $\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ — add the two copies when $f(x) + f(a-x)$ simplifies.
4. Even/odd on symmetric intervals: $\int_{-a}^{a} f = 2\int_0^a f$ for even $f$, and $= 0$ for odd $f$ — check the parity before integrating; many scary integrals are odd and die instantly.

The king-property template: for $I = \int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx$, replacing $x$ by $\frac{\pi}{2} - x$ swaps sine and cosine, and adding the two versions gives $2I = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}$, so $I = \frac{\pi}{4}$ — the single most-recycled definite integral on board papers.

Worked Examples — Indefinite Integrals

Example 1 — Table warm-up

Evaluate $\int\left(2x + 3\cos x + e^x\right)dx$.

Step 1 — term by term: $x^2 + 3\sin x + e^x$.

Answer: $x^2 + 3\sin x + e^x + C$ — one $C$ for the whole integral, never one per term.

Example 2 — Deriving $\int\tan x\,dx$

Evaluate $\int \tan x\,dx$.

Step 1 — write as a quotient: $\int\frac{\sin x}{\cos x}dx$; put $t = \cos x$, $dt = -\sin x\,dx$.

Step 2 — integrate: $-\int\frac{dt}{t} = -\log\vert t\vert = -\log\vert\cos x\vert$.

Answer: $\log\vert\sec x\vert + C$ — the two forms $-\log\vert\cos x\vert$ and $\log\vert\sec x\vert$ are the same answer.

Example 3 — The $\frac{f'}{f}$ shape

Evaluate $\int\dfrac{2x}{1 + x^2}dx$.

Step 1 — recognise numerator = derivative of denominator: so the answer is the log of the denominator.

Answer: $\log(1 + x^2) + C$ (no modulus needed — the denominator is always positive).

Example 4 — Substitution with a power inside

Evaluate $\int x\sin(x^2)\,dx$.

Step 1 — put $t = x^2$: $dt = 2x\,dx$, so the integral is $\frac{1}{2}\int\sin t\,dt$.

Step 2 — integrate and return: $-\frac{1}{2}\cos t$.

Answer: $-\dfrac{1}{2}\cos(x^2) + C$ — the visible factor $x$ is what makes the substitution work.

Example 5 — Completing the square

Evaluate $\int\dfrac{dx}{x^2 + 2x + 2}$.

Step 1 — complete the square: $x^2 + 2x + 2 = (x + 1)^2 + 1$.

Step 2 — match the $\tan^{-1}$ form with $a = 1$:

Answer: $\tan^{-1}(x + 1) + C$ — every irreducible quadratic denominator routes through this two-step.

Example 6 — By parts, ILATE in action

Evaluate $\int x\cos x\,dx$.

Step 1 — choose $u$ by ILATE: Algebraic $x$ beats Trigonometric $\cos x$, so $u = x$, $v = \cos x$.

Step 2 — apply the formula: $x\sin x - \int 1\cdot\sin x\,dx = x\sin x + \cos x$.

Answer: $x\sin x + \cos x + C$ — check by differentiating: $\sin x + x\cos x - \sin x = x\cos x$. ✓

Example 7 — The famous $\int\log x\,dx$

Evaluate $\int\log x\,dx$.

Step 1 — take $u = \log x$, $v = 1$: $\log x \cdot x - \int\frac{1}{x}\cdot x\,dx$.

Step 2 — finish: $x\log x - x$.

Answer: $x\log x - x + C$ — the invisible factor $1$ is the trick worth remembering.

Example 8 — The $e^x(f + f')$ shortcut

Evaluate $\int e^x(\sin x + \cos x)\,dx$.

Step 1 — match the template: with $f(x) = \sin x$, $f'(x) = \cos x$ — exactly $e^x[f + f']$.

Answer: $e^x\sin x + C$ — no by-parts loop needed once the shape is spotted.

Example 9 — Algebraic times exponential

Evaluate $\int xe^x\,dx$.

Step 1 — by parts with $u = x$: $xe^x - \int e^x dx = xe^x - e^x$.

Answer: $(x - 1)e^x + C$.

Example 10 — Partial fractions

Evaluate $\int\dfrac{dx}{(x+1)(x+2)}$.

Step 1 — split: $\frac{1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$; covering up gives $A = 1$, $B = -1$.

Step 2 — integrate term by term: $\log\vert x+1\vert - \log\vert x+2\vert$.

Answer: $\log\left\vert\dfrac{x+1}{x+2}\right\vert + C$.

Worked Examples — Definite Integrals

Example 11 — Fundamental theorem

Evaluate $\int_0^1 x^2\,dx$.

Step 1 — antiderivative and limits: $\left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3} - 0$.

Answer: $\dfrac{1}{3}$ — no $+C$ in a definite answer.

Example 12 — A trigonometric definite

Evaluate $\int_0^{\pi/4}\sec^2 x\,dx$.

Step 1 — antiderivative: $\left[\tan x\right]_0^{\pi/4} = 1 - 0$.

Answer: $1$.

Example 13 — The king property classic

Evaluate $I = \int_0^{\pi/2}\dfrac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx$.

Step 1 — apply $x \to \frac{\pi}{2} - x$: $I = \int_0^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}dx$.

Step 2 — add the two versions: the fractions are complementary: $$2I = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}$$

Answer: $I = \dfrac{\pi}{4}$ — the add-and-collapse works for every complementary pair on $\left[0, \frac{\pi}{2}\right]$.

Example 14 — Parity kills an integral

Evaluate $\int_{-1}^{1} x^3\,dx$ and state the general principle.

Step 1 — check parity: $x^3$ is odd ($f(-x) = -f(x)$) and the interval is symmetric.

Answer: $0$ by the odd-function property — and for an even integrand the same symmetry gives $\int_{-a}^{a} = 2\int_0^a$. Check parity before computing anything.

Example 15 — King property with squares

Evaluate $I = \int_0^{\pi/2}\sin^2 x\,dx$.

Step 1 — king: $I = \int_0^{\pi/2}\sin^2\left(\frac{\pi}{2} - x\right)dx = \int_0^{\pi/2}\cos^2 x\,dx$.

Step 2 — add: $2I = \int_0^{\pi/2}(\sin^2 x + \cos^2 x)\,dx = \frac{\pi}{2}$.

Answer: $I = \dfrac{\pi}{4}$ — same collapse as Example 13, and faster than the double-angle route.