Evaluate:∫tan−1xdx
11+x2
xtan-1x+12log|1+x2|
xtan-1x+12tan-1x1+x2
xtan-1x-12log|1+x2|
Evaluating the integral∫tan−1xdx
⇒∫tan−1x1dx⇒tan−1x∫1dx−∫ddxtan−1x∫1dxdxApplyingintegrationbyparts⇒xtan−1x−∫x1+x2dx
Substituting 1+x2=tand put the value of dxafter differentiating w.r.t. x
dx=dt2x
⇒xtan-1x-12∫1tdt+c⇒xtan-1x-12log|t|+c⇒xtan-1x-12log|1+x2|+c[substitutingt=(1+x2)]
Hence, option D is the correct answer.