What is \(\rm \int^1_0 \frac {\tan^{-1} x}{1 + x^2}dx\) equal to?

Question

Question

Answer 1

Concept:

\( \rm \int x^{n}dx = \frac{x^{n+1}}{n+1} + c\)

Calculation:

Let I = \(\rm \int^1_0 \frac {\tan^{-1} x}{1 + x^2}dx\)

Let tan^-1 x = t

Differentiating both sides, we get

⇒ \(\rm \frac{1}{1+x^2} dx = dt\)

x	0	1
t	0	π/4

Now, I \(= \rm \int _0^{\pi/4} t\;dt\)

\( = \rm [\frac{t^2}{2}]_0^{\pi/4}\)

\(= \rm \frac{1}{2}[\frac{\pi^2}{16}-0]\)

= \(\rm \frac {\pi^2}{32}\)