Evaluate the integral of (tan^-1 x)/(1+x² )dx from 0 to1
Let u=tan−1xu=tan^{-1}xu=tan−1x
Then dudx=11+x2{du \over dx}={1 \over 1+x^2}dxdu=1+x21
dx=(1+x2)dudx=(1+x²)dudx=(1+x2)du
∫0u1+x2.(1+x2)du\int_0 {u \over 1+x²}.(1+x²)du∫01+x2u.(1+x2)du
=u22=(tan−1x)22{u² \over 2}= {(tan^{-1}x)^2 \over 2}2u2=2(tan−1x)2 From 0 to 1
(tan−11)22−(tan−10)22{(tan^{-1}1)² \over 2}-{(tan^{-1}0)² \over 2}2(tan−11)2−2(tan−10)2
(tan−11)22{(tan^{-1}1)² \over 2}2(tan−11)2
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