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.

Trigonometric Integrals

Even powers — reduce the power first: use $\sin^2 x = \frac{1 - \cos 2x}{2}$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$. Never integrate an even power directly — convert to first-degree cosines of multiple angles, then use the table.

Odd powers — peel one factor and substitute: write $\sin^3 x = \sin x\left(1 - \cos^2 x\right)$ and put $t = \cos x$. The peeled factor becomes the $dt$; what remains is a polynomial in $t$.

Products with different angles — product-to-sum identities: $$2\sin A\cos B = \sin(A+B) + \sin(A-B)$$ $$2\cos A\cos B = \cos(A+B) + \cos(A-B), \qquad 2\sin A\sin B = \cos(A-B) - \cos(A+B)$$ Convert the product into a sum of single sines and cosines, then integrate term by term.

The $(px + q)$ Numerator Technique

For $\int\frac{px+q}{ax^2+bx+c}\,dx$ or $\int\frac{px+q}{\sqrt{ax^2+bx+c}}\,dx$, express the numerator through the derivative of the quadratic: $$px + q = A\,(2ax + b) + B$$ and compare coefficients to find $A$ and $B$. The $A$-part is then an $\frac{f'}{f}$ logarithm (or $2\sqrt{f}$ when the quadratic sits under a root), and the $B$-part is a standard special form after completing the square. This two-line split is the most heavily tested indefinite-integration move on board papers.

The Square-Root Integrals (learn all three)

$$\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$$ $$\int\sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log\left\vert x + \sqrt{x^2 + a^2}\right\vert + C$$ $$\int\sqrt{x^2 - a^2}\,dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\log\left\vert x + \sqrt{x^2 - a^2}\right\vert + C$$ Memory hook: the first term is always $\frac{x}{2}$ times the root itself; the second carries $\frac{a^2}{2}$ and copies the matching $\int\frac{dx}{\sqrt{\;\cdot\;}}$ partner — $\sin^{-1}$ for $a^2 - x^2$, the log for the other two, with a minus sign only in the $x^2 - a^2$ case. For a general quadratic under the root, complete the square first and then apply the matching result.

Partial Fractions — The Full Menu

Factor in the denominator	Terms you must write
$(x - a)$	$\dfrac{A}{x - a}$
$(x - a)^2$	$\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$
$x^2 + bx + c$ (irreducible)	$\dfrac{Ax + B}{x^2 + bx + c}$

A repeated factor needs both terms — writing only $\frac{B}{(x-a)^2}$ is the classic half-mark error.

Improper fractions — divide first. When the numerator's degree is at least the denominator's, do the polynomial division before splitting: $\frac{x^2}{x^2 + 1} = 1 - \frac{1}{1 + x^2}$. One line of division saves a page of wrong algebra.

Definite Integrals — Substitution and the $2a$ Property

Substitution in a definite integral: change the limits and never come back to $x$. In $\int_0^1\frac{x}{x^2+1}\,dx$ with $t = x^2 + 1$, the limits move to $t = 1$ and $t = 2$, and the answer is read off directly in $t$.

The $2a$ property: $$\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a - x)\,dx$$ so the integral doubles when $f(2a - x) = f(x)$ and vanishes when $f(2a - x) = -f(x)$. Combined with the king's property on $[0, \pi]$, this cracks the celebrated $\int_0^{\pi}\frac{x\sin x}{1 + \cos^2 x}\,dx$ family — see Example 26.

Worked Examples — Trigonometric Integrals

Example 16 — Even power of sine

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

Step 1 — power-reduce: $\sin^2 x = \dfrac{1 - \cos 2x}{2}$.

Step 2 — integrate term by term: $\dfrac{x}{2} - \dfrac{\sin 2x}{4}$.

Answer: $\dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ — the identity does the work; the table finishes it.

Example 17 — Product to sum

Evaluate $\int\sin 3x\cos 4x\,dx$.

Step 1 — convert the product: $2\sin 3x\cos 4x = \sin 7x + \sin(-x)$, so $\sin 3x\cos 4x = \dfrac{1}{2}\left(\sin 7x - \sin x\right)$.

Step 2 — integrate: $\dfrac{1}{2}\left(-\dfrac{\cos 7x}{7} + \cos x\right)$.

Answer: $-\dfrac{\cos 7x}{14} + \dfrac{\cos x}{2} + C$ — watch the sign: $\sin(-x) = -\sin x$.

Example 18 — Odd power of sine

Evaluate $\int\sin^3 x\,dx$.

Step 1 — peel one sine: $\int\sin x\left(1 - \cos^2 x\right)dx$; put $t = \cos x$, $dt = -\sin x\,dx$.

Step 2 — polynomial in $t$: $-\int\left(1 - t^2\right)dt = -t + \dfrac{t^3}{3}$.

Answer: $-\cos x + \dfrac{\cos^3 x}{3} + C$.

Worked Examples — The $(px+q)$ Split and Root Forms

Example 19 — Linear over quadratic

Evaluate $\int\dfrac{x + 3}{x^2 + 2x + 5}\,dx$.

Step 1 — split the numerator through the derivative: $x + 3 = \dfrac{1}{2}(2x + 2) + 2$.

Step 2 — the $\frac{f'}{f}$ part: $\dfrac{1}{2}\int\dfrac{2x+2}{x^2+2x+5}dx = \dfrac{1}{2}\log\left(x^2 + 2x + 5\right)$.

Step 3 — the constant part, complete the square: $x^2 + 2x + 5 = (x+1)^2 + 4$, so $2\int\dfrac{dx}{(x+1)^2 + 2^2} = 2\cdot\dfrac{1}{2}\tan^{-1}\dfrac{x+1}{2}$.

Answer: $\dfrac{1}{2}\log\left(x^2 + 2x + 5\right) + \tan^{-1}\dfrac{x+1}{2} + C$ — the two-part answer (log plus inverse tan) is the signature of this technique.

Example 20 — Root of a quadratic

Evaluate $\int\dfrac{dx}{\sqrt{5 - 4x - x^2}}$.

Step 1 — complete the square inside the root: $5 - 4x - x^2 = 9 - (x + 2)^2$.

Step 2 — match $\int\frac{dx}{\sqrt{a^2 - u^2}} = \sin^{-1}\frac{u}{a}$ with $a = 3$, $u = x + 2$:

Answer: $\sin^{-1}\dfrac{x + 2}{3} + C$.

Example 21 — The $\sqrt{a^2 - x^2}$ result in action

Evaluate $\int\sqrt{4 - x^2}\,dx$.

Step 1 — identify $a = 2$ and quote the standard result:

Answer: $\dfrac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\dfrac{x}{2} + C$ — the coefficient of the inverse sine is $\frac{a^2}{2} = 2$, not $a$.

Worked Examples — Partial Fractions, All Cases

Example 22 — Repeated linear factor

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

Step 1 — write all three terms: $\dfrac{1}{(x-1)^2(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{D}{x+2}$; cover-up at $x = 1$ gives $B = \dfrac{1}{3}$, at $x = -2$ gives $D = \dfrac{1}{9}$, and comparing constants gives $A = -\dfrac{1}{9}$.

Step 2 — integrate each: $-\dfrac{1}{9}\log\vert x-1\vert - \dfrac{1}{3(x-1)} + \dfrac{1}{9}\log\vert x+2\vert$.

Answer: $\dfrac{1}{9}\log\left\vert\dfrac{x+2}{x-1}\right\vert - \dfrac{1}{3(x-1)} + C$ — note the $(x-1)^{-2}$ term integrates to an algebraic term, not a log.

Example 23 — Irreducible quadratic factor

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

Step 1 — split with a linear numerator on the quadratic: $\dfrac{1}{x(x^2+1)} = \dfrac{1}{x} - \dfrac{x}{x^2+1}$.

Step 2 — integrate: $\log\vert x\vert - \dfrac{1}{2}\log\left(x^2 + 1\right)$.

Answer: $\log\vert x\vert - \dfrac{1}{2}\log\left(x^2 + 1\right) + C$ — the second term is an $\frac{f'}{f}$ log after adjusting by $\frac{1}{2}$.

Example 24 — Improper fraction: divide first

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

Step 1 — divide: $\dfrac{x^2}{x^2+1} = 1 - \dfrac{1}{1 + x^2}$.

Answer: $x - \tan^{-1}x + C$ — partial fractions without dividing first would have failed silently.

Worked Examples — Definite Techniques

Example 25 — Substitution with limit change

Evaluate $\int_0^1\dfrac{x}{x^2 + 1}\,dx$.

Step 1 — put $t = x^2 + 1$, $dt = 2x\,dx$, and move the limits: $x = 0 \Rightarrow t = 1$; $x = 1 \Rightarrow t = 2$.

Step 2 — evaluate in $t$: $\dfrac{1}{2}\int_1^2\dfrac{dt}{t} = \dfrac{1}{2}\left[\log t\right]_1^2$.

Answer: $\dfrac{1}{2}\log 2$ — no return to $x$, no constant $C$.

Example 26 — The $\frac{\pi^2}{4}$ classic

Evaluate $I = \int_0^{\pi}\dfrac{x\sin x}{1 + \cos^2 x}\,dx$.

Step 1 — king's property with $a = \pi$: $I = \int_0^{\pi}\dfrac{(\pi - x)\sin x}{1 + \cos^2 x}\,dx$ (sine and cosine-squared are unchanged under $x \to \pi - x$).

Step 2 — add the two versions: $2I = \pi\int_0^{\pi}\dfrac{\sin x}{1 + \cos^2 x}\,dx$.

Step 3 — substitute $t = \cos x$, limits $1$ to $-1$: $2I = \pi\int_{-1}^{1}\dfrac{dt}{1 + t^2} = \pi\left[\tan^{-1}t\right]_{-1}^{1} = \pi\cdot\dfrac{\pi}{2}$.

Answer: $I = \dfrac{\pi^2}{4}$ — king's property removes the $x$, substitution finishes the rest; the single most famous definite integral in the chapter.