What is the integration of tan inverse root x ?

Solution :

Let I = \(\int\) \(tan^{-1}\sqrt{x}\).1 dx

By Applying integration by parts,

Taking \(tan^{-1}\sqrt{x}\) as first function and 1 as second function. Then

I = \(tan^{-1}\sqrt{x}\) \(\int\) 1 dx – \(\int\) {\(d\over dx\)\(tan^{-1}\sqrt{x}\) \(\int\) 1 dx } dx

I = x\(tan^{-1}\sqrt{x}\) – \(\int\) \(1\over 2(1+x)\sqrt{x}\) . x dx

Let \(\sqrt{x}\) = t

\(1\over 2\sqrt{x}\) dx = dt \(\implies\) dx = 2t dt

\(\implies\) I = x\(tan^{-1}\sqrt{x}\) – \(\int\) \(t^2\over t^2 + 1\) dt

\(\implies\) I = x\(tan^{-1}\sqrt{x}\) – \(\int\) dt + \(\int\) \(1\over 1 + t^2\)

\(\implies\) I = x\(tan^{-1}\sqrt{x}\) – t + \(tan^{-1}t\) + C

Hence, I = x\(tan^{-1}\sqrt{x}\) – \(\sqrt{x}\) + \(tan^{-1}\sqrt{x}\) + C