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Question

Question: \(\int _ { - 1 } ^ { 1 } \frac { 1 } { 1 + x ^ { 2 } }\) =...

∫−1111+x2\int _ { - 1 } ^ { 1 } \frac { 1 } { 1 + x ^ { 2 } } =

A

p

B

π2\frac { \pi } { 2 }

C

– π2\frac { \pi } { 2 }

D

π3\frac { \pi } { 3 }

Answer

π2\frac { \pi } { 2 }

Explanation

Solution

[tan⁡−1x]−11\left[ \tan ^ { - 1 } x \right] _ { - 1 } ^ { 1 } =

tan–1(1)–tan–1(–1)=π4\frac { \pi } { 4 }–(−π4)\left( - \frac { \pi } { 4 } \right)= π2\frac { \pi } { 2 }