Browse Questions

# Integrate the function$\tan^{-1}x$

$\begin{array}{1 1} x\tan^{-1} x- \frac{1}{2} \log|1+x^2|+c \\ x\tan^{-1}x+ \frac{1}{2} \log|1+x^2|+c \\ x\tan^{-1}x- \frac{1}{2} \log|1-x^2|+c \\ x\tan^{-1}x+ \frac{1}{2} \log|1-x^2|+c \end{array}$

Toolbox:
• (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts$\int udv=uv-\int vdu$
• (ii)$\;\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^2}.$
• (iii)Method of substitution :If f(x)=t,then f'(x)dx=dt.
• Hence $\int f(x)=\int t.dt+c.$
Given $I=\int \tan^{-1}xdx.$

Clearly the given integral function is of the form $\int u dv$

Let $u=\tan^{-1}x.$Hence on differentiating we get,

du=$\frac{1}{1+x^2}dx.$

Let dv=dx,on integrating we get

v=x.

Now substituting for u,v,du and dv we get,

$\int \tan^{-1}xdx=x\tan^{-1}x-\int x.\frac{1}{1+x^2}dx.$

Let $I_1=\frac{x}{1+x^2}dx.$

Put $1+x^2=t.$

On differentiating with respect to x,

2xdx=dt.

xdx=$\frac{dt}{2}.$

Now substituting for t and dt we get,

$\frac{dt/2}{t}=\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}log|t|.$

Substituting for t we get

$\qquad=\frac{1}{2}log|1+x^2|.$

Now substituting for $I_1$ we get $I=x\tan^{-1}x-\frac{1}{2}log|1+x^2|+c.$