$\displaystyle\int_0^{1} \dfrac{\tan^{-1}x}{1+x^2}\,dx$ equals
A$\dfrac{\pi^2}{16}$
B$\dfrac{\pi^2}{32}$
C$\dfrac{\pi}{4}$
D$\dfrac{\pi^2}{8}$
Answer & Solution
Correct answer: B. $\dfrac{\pi^2}{32}$
1. Put $t=\tan^{-1}x$, so $dt=\dfrac{dx}{1+x^2}$.
2. Change limits: $x=0\Rightarrow t=0$; $x=1\Rightarrow t=\tfrac{\pi}{4}$.
3. Integral $=\int_0^{\pi/4} t\,dt=\left[\dfrac{t^2}{2}\right]_0^{\pi/4}=\dfrac{1}{2}\cdot\dfrac{\pi^2}{16}$.
4. $=\dfrac{\pi^2}{32}$. Option A forgets the factor $\tfrac12$.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.341_
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