$\displaystyle\int_0^{1} \dfrac{ an^{-1}x}{1+x^2}\,dx$ equals

Question

$\displaystyle\int_0^{1} \dfrac{	an^{-1}x}{1+x^2}\,dx$ equals

Accepted Answer

B. $\dfrac{\pi^2}{32}$

1. Put $t=	an^{-1}x$, so $dt=\dfrac{dx}{1+x^2}$.
2. Change limits: $x=0\Rightarrow t=0$; $x=1\Rightarrow t=	frac{\pi}{4}$.
3. Integral $=\int_0^{\pi/4} t\,dt=\left[\dfrac{t^2}{2}ight]_0^{\pi/4}=\dfrac{1}{2}\cdot\dfrac{\pi^2}{16}$.
4. $=\dfrac{\pi^2}{32}$. Option A forgets the factor $	frac12$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.341_

Answer

$\dfrac{\pi^2}{16}$

Answer

$\dfrac{\pi}{4}$

Answer

$\dfrac{\pi^2}{8}$

$\displaystyle\int_0^{1} \dfrac{\tan^{-1}x}{1+x^2}\,dx$ equals

Keep practising Straight Lines and Conic Sections