Integrals
A function \(F\) is an antiderivative (primitive) of \(f\) if \(F\,'(x)=f(x)\). Since constants differentiate to zero, antiderivatives differ by constants, and we write the indefinite integral \(\int f(x)\,dx=F(x)+C\).
Every derivative formula, read backwards, gives an integral:
| \(f(x)\) | \(\int f(x)\,dx\) |
|---|---|
| \(x^n\ (n\neq-1)\) | \(\frac{x^{n+1}}{n+1}+C\) |
| \(\frac1x\) | \(\log|x|+C\) |
| \(e^x\) | \(e^x+C\) |
| \(a^x\) | \(\frac{a^x}{\log a}+C\) |
| \(\sin x\) | \(-\cos x+C\) |
| \(\cos x\) | \(\sin x+C\) |
| \(\sec^2x\) | \(\tan x+C\) |
| \(\csc^2x\) | \(-\cot x+C\) |
| \(\sec x\tan x\) | \(\sec x+C\) |
| \(\frac{1}{\sqrt{1-x^2}}\) | \(\sin^{-1}x+C\) |
| \(\frac1{1+x^2}\) | \(\tan^{-1}x+C\) |
Integration is linear: \(\int[kf(x)\pm g(x)]\,dx=k\int f\,dx\pm\int g\,dx\). Rewrite roots and fractions as powers first: \(\int\frac{x^2+1}{\sqrt x}\,dx=\int(x^{3/2}+x^{-1/2})\,dx=\frac25x^{5/2}+2x^{1/2}+C\).
To find a particular antiderivative, use a given condition to fix \(C\): if \(F\,'(x)=3x^2\) and \(F(1)=4\), then \(F(x)=x^3+3\).
| \(f(x)\) | \(\int f(x)\,dx\) |
|---|---|
| \(x^n\ (n\neq-1)\) | \(\frac{x^{n+1}}{n+1}+C\) |
| \(\frac1x\) | \(\log|x|+C\) |
| \(e^x\) | \(e^x+C\) |
| \(a^x\) | \(\frac{a^x}{\log a}+C\) |
| \(\sin x\) | \(-\cos x+C\) |
| \(\cos x\) | \(\sin x+C\) |
| \(\sec^2x\) | \(\tan x+C\) |
| \(\csc^2x\) | \(-\cot x+C\) |
| \(\sec x\tan x\) | \(\sec x+C\) |
| \(\frac{1}{\sqrt{1-x^2}}\) | \(\sin^{-1}x+C\) |
| \(\frac1{1+x^2}\) | \(\tan^{-1}x+C\) |
\(\int x\sin x\,dx\neq\int x\,dx\cdot\int\sin x\,dx\). Products need substitution or by parts.
\(\int x^{-1}dx=\log|x|+C\), but \(\int x^{-2}dx=-x^{-1}+C\) – the log rule is only for the power −1.
A fresh set of questions from this topic. Change answers freely – solutions appear after you submit.
Start PracticeIntegration undoes differentiation; learn the table, use linearity and simplify the integrand into standard forms.
Create a free account to take practice sessions and tests, see detailed explanations and track your progress.
