Evaluate _ 0 ^ 1 ^ - 1 x 1 + x ^ 2 dx - Vidyadip

Question

Evaluate $\int _ { 0 } ^ { 1 } \frac { \tan ^ { - 1 } x } { 1 + x ^ { 2 } } dx$

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Answer

Let $I = \int _ { 0 } ^ { 1 } \frac { \tan ^ { - 1 } x } { 1 + x ^ { 2 } } d x$
Put $\tan ^ { - 1 } x = t \quad \Rightarrow \frac { 1 } { 1 + x ^ { 2 } } d x = d t$
Lower limit when x = 0, then t = 0
Upper limit when x = 1, then t = $\pi / 4.$
$\therefore I = \int _ { 0 } ^ { \pi / 4 } t d t = \left[ \frac { t ^ { 2 } } { 2 } \right] _ { 0 } ^ { \pi / 4 } = \frac { 1 } { 2 } \left[ \left( \frac { \pi } { 4 } \right) ^ { 2 } - ( 0 ) ^ { 2 } \right] = \frac { \pi ^ { 2 } } { 32 }$

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