Basic Rules of Integration

Because integration is defined as the reverse of differentiation, every standard derivative formula immediately hands us a matching standard antiderivative — simply read the differentiation rule backwards. Collecting them gives the working toolkit for this whole chapter.

Each row is literally the mirror image of an earlier derivative fact: since $\dfrac{d}{dx}(c)=0$, we get $\int 0\,dx=c$; since $\dfrac{d}{dx}(kx)=k$, we get $\int k\,dx=kx+c$; since $\dfrac{d}{dx}\!\left(\dfrac{x^{n+1}}{n+1}\right)=x^n$, we get the Power rule (the restriction $n\ne-1$ is forced because $n+1=0$ would divide by zero — the $n=-1$ case is instead covered separately by the $\log|x|$ row, coming from $\dfrac{d}{dx}(\log x)=\dfrac1x$); and so on down every trigonometric, exponential and inverse-trigonometric row.

Applying the power rule directly. Any integrand that can be rewritten as a single power of $x$ (including negative and fractional powers, via $\dfrac{1}{x^n}=x^{-n}$ and $\sqrt[k]{x^n}=x^{n/k}$) integrates by the power rule after that rewriting. For instance $\displaystyle\int x^{10}\,dx = \frac{x^{11}}{11}+c$; and reciprocal powers integrate the same way after converting to a negative exponent, e.g. $\displaystyle\int \frac{1}{x^{10}}\,dx=\int x^{-10}\,dx=\frac{x^{-9}}{-9}+c=-\frac{1}{9x^9}+c$. Roots integrate identically once written as fractional powers: $\displaystyle\int \sqrt{x}\,dx=\int x^{1/2}\,dx=\frac{x^{3/2}}{3/2}+c=\tfrac23 x^{3/2}+c$, and $\displaystyle\int \frac{1}{\sqrt x}\,dx=\int x^{-1/2}\,dx=\frac{x^{1/2}}{1/2}+c=2\sqrt x+c$.

Reducing a trig quotient to a standard form. Many integrands that don't look like they are on the table above are simply a standard trig antiderivative in disguise. Recognising a familiar identity is the whole trick:

• $\dfrac{1}{\cos^2 x}=\sec^2 x$, so $\displaystyle\int \frac{dx}{\cos^2 x}=\int\sec^2x\,dx=\tan x+c$.
• $\dfrac{\cot x}{\sin x}=\dfrac{\cos x}{\sin^2 x}=\text{cosec}\,x\cot x$, so $\displaystyle\int \frac{\cot x}{\sin x}\,dx=-\text{cosec}\,x+c$.
• $\dfrac{\sin x}{\cos^2 x}=\dfrac{\sin x}{\cos x}\cdot\dfrac{1}{\cos x}=\tan x\sec x$, so $\displaystyle\int \frac{\sin x}{\cos^2 x}\,dx=\int\tan x\sec x\,dx=\sec x+c$.
• $\dfrac{1}{\sqrt{1-x^2}}$ and $\dfrac{1}{1+x^2}$ integrate directly to $\sin^{-1}x$ and $\tan^{-1}x$ respectively, straight off the table. …