MCQ
$\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x=$
- A$\frac{\pi^2}{8}$
- B$\frac{\pi^2}{16}$
- C$\frac{\pi^2}{4}$
- ✓$\frac{\pi^2}{32}$
✓
Answer
Correct option: D.
$\frac{\pi^2}{32}$
(D)
Put $\tan ^{-1} x= t \Rightarrow \frac{1}{1+x^2} d x= dt$
When $x-0, t -0$ and when $x-1, t -\frac{\pi}{4}$
$\therefore \quad \int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x=\int_0^{\pi / 4} tdt =\left[\frac{ t ^2}{2}\right]_0^{\frac{\pi}{4}}=\frac{\pi^2}{32}$
Put $\tan ^{-1} x= t \Rightarrow \frac{1}{1+x^2} d x= dt$
When $x-0, t -0$ and when $x-1, t -\frac{\pi}{4}$
$\therefore \quad \int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x=\int_0^{\pi / 4} tdt =\left[\frac{ t ^2}{2}\right]_0^{\frac{\pi}{4}}=\frac{\pi^2}{32}$
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