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Question

Question: \(\int _ { 0 } ^ { 1 } \tan ^ { - 1 } \left( \frac { 1 } { x ^ { 2 } - x + 1 } \right)\) dx is...

∫01tan⁡−1(1x2−x+1)\int _ { 0 } ^ { 1 } \tan ^ { - 1 } \left( \frac { 1 } { x ^ { 2 } - x + 1 } \right) dx is

A

ln 2

B

– ln 2

C

π2\frac { \pi } { 2 } + ln 2

D

π2\frac { \pi } { 2 } – ln 2

Answer

π2\frac { \pi } { 2 } – ln 2

Explanation

Solution

∫01tan⁡−111+x(x−1)⋅dx\int _ { 0 } ^ { 1 } \tan ^ { - 1 } \frac { 1 } { 1 + x ( x - 1 ) } \cdot d x

=

= ∫01(tan⁡−1x−tan⁡−1(x−1))dx\int _ { 0 } ^ { 1 } \left( \tan ^ { - 1 } x - \tan ^ { - 1 } ( x - 1 ) \right) d x

= [P-4]

= 2