Which integration technique do I use?
Match the integrand's shape to the technique built for it.
Look at the integrand's overall shape before picking a technique. A product of a polynomial and a transcendental function (exponential, log, trig, inverse trig) wants integration by parts — repeated if the polynomial factor needs differentiating more than once, via tabular integration. A rational function with a factorable denominator wants partial fractions. A square root of a^2\pm x^2 or x^2-a^2 wants a trig substitution (completing the square first if the quadratic isn't already centered). A composite function whose inner derivative also appears wants u-substitution — try this FIRST, since it's the fastest when it works.
Substitution-based
Formula
u-Substitution
If u = g(x) with g differentiable and f continuous on the range of g, then \int f(g(x))\,g'(x)\,dx = \int f(u)\,du. For a definite integral, the bounds transform along with the variable: \int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.
| Radical present | Substitute | Identity used | Radical becomes |
|---|---|---|---|
| \sqrt{a^2-x^2} | x=a\sin\theta | 1-\sin^2\theta=\cos^2\theta | a\cos\theta |
| \sqrt{a^2+x^2} | x=a\tan\theta | 1+\tan^2\theta=\sec^2\theta | a\sec\theta |
| \sqrt{x^2-a^2} | x=a\sec\theta | \sec^2\theta-1=\tan^2\theta | a\tan\theta |
In every row, dx is replaced using the derivative of the substitution (e.g. dx = a\cos\theta\,d\theta in the first row), and once the integral is evaluated in \theta, a right triangle built from the substitution equation converts the result back to a function of x.
Example
For an integrand built from \sin x and \cos x combined only by addition, subtraction, multiplication, and division (a rational function of \sin x and \cos x), the substitution t=\tan\!\left(\dfrac{x}{2}\right) converts the ENTIRE integral into a rational function of t: \sin x = \frac{2t}{1+t^2}, \qquad \cos x = \frac{1-t^2}{1+t^2}, \qquad dx = \frac{2}{1+t^2}\,dt.
The universal fallback for a rational function of sin and cos, when no other technique applies cleanly.
Parts-based
For differentiable u=u(x) and v=v(x): \int u\,dv = uv - \int v\,du.
Show proof
Start from the product rule, \dfrac{d}{dx}[uv] = u\,\dfrac{dv}{dx} + v\,\dfrac{du}{dx}, and integrate both sides with respect to x: uv = \int u\,\frac{dv}{dx}\,dx + \int v\,\frac{du}{dx}\,dx = \int u\,dv + \int v\,du. Solving for \int u\,dv gives the formula directly. No information is added or lost — it's the product rule read from right to left, with the two terms it produces separated onto opposite sides of the equation.
When u is a polynomial (or anything that reaches zero after finitely many derivatives) and dv is easy to integrate repeatedly, build two columns: D (successive derivatives of u, ending at 0) and I (successive antiderivatives of dv). Read the answer off as the diagonal products, with signs alternating +,-,+,-,\dots, stopping once the D-column hits 0: \int u\,dv = u_0 I_1 - u_1 I_2 + u_2 I_3 - \cdots \pm u_n I_{n+1} + C.
Use when u must be differentiated more than once (e.g. a polynomial factor of degree 2+).
Show proof
Each row is one honest application of integration by parts: \int u\,dv = uv - \int v\,du moves one derivative onto u and one antiderivative onto dv, exactly matching the next D/I row. Repeating this on the leftover integral \int v\,du each time is what produces the alternating sign — every fresh application contributes another minus sign in front of its own leftover integral. The process terminates because u's derivative column reaches 0 after finitely many steps whenever u is a polynomial, at which point the "leftover integral" is \int 0\cdot(\text{anything})\,dx = 0 and nothing remains to integrate by parts again — the table's diagonal products are the whole answer.
Rational functions
For q(x) = (x - r_1)(x - r_2)\cdots(x - r_n) with the r_i all distinct: \frac{p(x)}{q(x)} = \frac{A_1}{x - r_1} + \frac{A_2}{x - r_2} + \cdots + \frac{A_n}{x - r_n}
If (x-r)^k divides q(x) and no higher power of (x-r) does, its contribution to the decomposition of \frac{p(x)}{q(x)} is \frac{A_1}{x-r} + \frac{A_2}{(x-r)^2} + \cdots + \frac{A_k}{(x-r)^k}.
Show proof
Write q(x) = (x-r)^k m(x) with m(r) \neq 0 (every other factor of q). The cover-up value A_1 = \dfrac{p(r)}{m(r)} is well-defined since m(r)\neq0, and \frac{p(x)}{q(x)} - \frac{A_1}{(x-r)^k} = \frac{p(x) - A_1\,m(x)}{(x-r)^k m(x)}. The new numerator vanishes at x=r, since p(r) - A_1 m(r) = p(r) - p(r) = 0 by exactly how A_1 was chosen — so by the Factor Theorem, (x-r) divides it: p(x) - A_1 m(x) = (x-r)\,p_1(x) for some polynomial p_1. Substituting back, \frac{p(x)}{q(x)} - \frac{A_1}{(x-r)^k} = \frac{(x-r)\,p_1(x)}{(x-r)^k m(x)} = \frac{p_1(x)}{(x-r)^{k-1}m(x)}, a rational function with the SAME shape as the one just started with, except the power of (x-r) in the denominator dropped from k to k-1. Repeat the identical step on \frac{p_1(x)}{(x-r)^{k-1}m(x)} to peel off A_2 and drop to k-2, and so on. After k repetitions every power of (x-r) has been stripped out, leaving exactly \frac{A_1}{x-r}+\cdots+\frac{A_k}{(x-r)^k} pulled out and a remaining proper fraction over m(x) — the other factors of q(x), handled the same way by the rest of the Partial Fraction Decomposition Recipe.
For a distinct irreducible quadratic factor ax^2+bx+c (with b^2-4ac<0): \frac{Ax+B}{ax^2+bx+c}
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