What is∫tan-1xdx ?
Integration of∫tan-1xdx:
Let
t=tan-1x⇒tant=x⇒sec2tdt=dx
Now, ∫tan-1xdx=∫tsec2tdt
Integrate using: ∫uvdx=u∫vdx-∫(dudx∫vdx)dx
∫tan-1xdx=∫tsec2tdt=ttant-∫dtdttantdt=ttant-∫tantdt[∵∫tantdt=lnsect]=ttant-logsect+c=ttant-log1+tan2t+c[∵sec2t=1+tan2t]=xtan-1x-log1+x2+c
Thus,the value of integration is∫tan-1xdx=xtan-1x-log|1+x2|+c .