Question

Solve ${\int }_0^{100}\left[{\tan }^{-1}\right(x\left)\right]dx$

Answer 1

$I={\int }_0^{100}{\tan }^{-1}xdx$

Using integration by parts,

$\int udv=uv-vdu$

Let $u={\tan }^{-1}x,dv=dx$

$du=\frac{1}{1+x^2},v=x$

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

$u_1=x^2+1$

$\Rightarrow \frac{d{u}_1}{dx}=2x$

$I=x{\tan }^{-1}x-\frac{1}{2}\int \frac{d{u}_1}{u_1}$

$=x{\tan }^{-1}x-\frac{1}{2}ln\left(u_1\right)+c$

$={x{\tan }^{-1}x-\frac{1}{2}ln(x^2+1)+c|}_0^{100}$

$=100{\tan }^{-1}\left(100\right)-\frac{1}{2}ln\left(10001\right)-0+0$

$=100{\tan }^{-1}\left(100\right)-\frac{1}{2}ln\left(10001\right)$.