Question
Evaluate the following integrals:
$\int\frac{1}{\sqrt{\tan^{-1}\text{x}}.(1+\text{x}^2)}\text{dx}$

Answer

$\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$\text{Let }\tan^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{1+\text{x}^2}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\text{Now,}\int\frac{\text{dx}}{\sqrt{\tan^{-1}\text{x}}(1+\text{x}^2)}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{\tan^{-1}\text{x}}+\text{C}$

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