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Q. $ \int tan^{-1} \sqrt{x}\,dx $ is equal to

Integrals

Solution:

We have, $I = \int 1\cdot tan^{-1}\sqrt{x}\,dx$

$ \Rightarrow I= tan^{-1}\sqrt{x} \cdot\left(x\right) -\int\frac{1}{ 1+x} \times \frac{1}{2\sqrt{x}} \times x \, dx $

$ = x\,tan^{-1}\sqrt{x}-\int\frac{x}{ \left(1+x\right) \,2\,\sqrt{x}} dx$

$ = x\, tan^{-1} \sqrt{x} -\int\left(\frac{1+x}{\left(1+x\right)\, 2\,\sqrt{x}} - \frac{1}{\left(1+x\right)\,2\,\sqrt{x}}\right) dx$

$= x\, tan^{-1} \sqrt{x}- \int\frac{dx}{2\,\sqrt{x}} + \int \frac{dx}{2\,\sqrt{x} \left(1+x\right)} $

$= x\,tan^{-1}\sqrt{x} - \sqrt{x} + tan^{-1}\sqrt{x} +C $

$ =\left(x+1\right)tan^{-1} \sqrt{x} - \sqrt{x} + C$