Integrate ∫tan-1xdx.
Apply Integration by parts:
Given, ∫tan-1xdx
Let, I=∫tan-1xdx
Let, t=tan-1x
⇒x=tant
Differentiating both sides with respect to t,
dxdt=sec2t⇒dx=sec2tdt⇒I=∫t·sec2dt
From integration by parts, we know that
∫f(x)·g(x)=f(x)∫g(x)dx-∫(dfxdx∫gxdx)dx
Here, take fx=t and gx=sec2t
⇒I=t∫sec2tdt-∫dtdt∫sec2tdtdt⇒I=t·tant-∫tantdt[∵∫sec2t=tant]⇒I=t·tant-lnsect+C[∵∫tant=lnsect]
Substituting back the value of t,
⇒I=tan-1x·x-lnsectan-1x+C⇒I=xtan-1x-ln1+x2+C[∵sectan-1x=1+x2,1+x2=1+x2]
Therefore, ∫tan-1(x)dx=xtan-1(x)-ln(1+x2)+C