Integration Rules

The table of basic integrals, the techniques for everything else, and the reason every indefinite integral ends in + C.

Why “+ C”?

Integration undoes differentiation. But differentiation throws away constants, since the derivative of 5 is 0, so undoing it can’t know which constant was there. Every antiderivative is really a whole family of curves stacked vertically:

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Every curve F(x)=x33−x+CF(x) = \tfrac{x^3}{3} - x + C has the same derivative, x2−1x^2 - 1. Drag across the graph: at any x, all the tangent lines are parallel. Sliding C moves the curve up and down without changing a single slope, so the derivative can’t tell them apart, and the integral must include + C+\,C.

Linearity

∫(af(x)+bg(x)) dx=a∫f(x) dx+b∫g(x) dx\int \big(a f(x) + b g(x)\big)\,dx = a\int f(x)\,dx + b\int g(x)\,dx. Constants come out and sums split, so polynomials integrate term by term.

Table of basic integrals

IntegralResult
∫k dx\int k\,dxkx+Ckx + C
∫xn dx\int x^n\,dxxn+1n+1+C(n≠−1)\frac{x^{n+1}}{n+1} + C\quad (n \ne -1)
∫1x dx\int \frac{1}{x}\,dxln⁡∣x∣+C\ln|x| + C
∫ex dx\int e^x\,dxex+Ce^x + C
∫ax dx\int a^x\,dxaxln⁡a+C\frac{a^x}{\ln a} + C
∫ln⁡x dx\int \ln x\,dxxln⁡x−x+Cx\ln x - x + C
∫sin⁡x dx\int \sin x\,dx−cos⁡x+C-\cos x + C
∫cos⁡x dx\int \cos x\,dxsin⁡x+C\sin x + C
∫sec⁡2x dx\int \sec^2 x\,dxtan⁡x+C\tan x + C
∫tan⁡x dx\int \tan x\,dx−ln⁡∣cos⁡x∣+C-\ln|\cos x| + C
∫sec⁡x dx\int \sec x\,dxln⁡∣sec⁡x+tan⁡x∣+C\ln|\sec x + \tan x| + C
∫11+x2 dx\int \frac{1}{1+x^2}\,dxarctan⁡x+C\arctan x + C
∫11−x2 dx\int \frac{1}{\sqrt{1-x^2}}\,dxarcsin⁡x+C\arcsin x + C

Techniques

u-substitution (the chain rule in reverse)

Look for a function and its derivative. In ∫2xcos⁡(x2) dx\int 2x\cos(x^2)\,dx, let u=x2u = x^2, du=2x dxdu = 2x\,dx: the integral becomes ∫cos⁡u du=sin⁡u+C=sin⁡(x2)+C\int \cos u\,du = \sin u + C = \sin(x^2) + C.

Integration by parts (the product rule in reverse)

∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du. For ∫xex dx\int x e^x\,dx take u=xu = x, dv=exdxdv = e^x dx: result xex−ex+Cxe^x - e^x + C. Choose uu by LIATE.

Partial fractions

Split a rational function into simple pieces: 1x2−1=1/2x−1−1/2x+1\frac{1}{x^2-1} = \frac{1/2}{x-1} - \frac{1/2}{x+1}, each of which integrates to a logarithm.

Trig identities

Reduce powers first: sin⁡2x=1−cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2}, so ∫sin⁡2x dx=x2−sin⁡2x4+C\int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C.

The integral calculator picks the technique automatically and shows each step. Single results with explanations: ∫ ln x, ∫ tan x, ∫ sec x, all integrals.

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Questions students ask

What are the basic integration rules?

Power rule ∫xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C (n≠−1n \ne -1), ∫1xdx=ln⁡∣x∣+C\int \frac1x dx = \ln|x| + C, ∫exdx=ex+C\int e^x dx = e^x + C, ∫sin⁡x dx=−cos⁡x+C\int \sin x\,dx = -\cos x + C, ∫cos⁡x dx=sin⁡x+C\int \cos x\,dx = \sin x + C, plus linearity: constants come out, and sums split.

Why is there no product rule for integrals?

The integral of a product is not the product of integrals. Instead, the product rule for derivatives turns into integration by parts: ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du.

How do I choose between substitution and parts?

If one factor is (a multiple of) the derivative of something inside another factor, use substitution. If you have a product of unrelated types (polynomial × exponential, polynomial × trig, log × anything), use parts, choosing uu with LIATE.

What is LIATE?

An order for choosing uu in integration by parts: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Pick the one that comes first as uu.