# $\int \tan^{-1}\sqrt{x}\: \: dx$ is equal to Option 1) $\left ( x+1 \right )\tan^{-1}\sqrt{x}-\sqrt{x}+C$ Option 2) $x\tan^{-1}\sqrt{x}-\sqrt{x}+C$ Option 3) $\sqrt{x}-xtan^{-1}\sqrt{x}+C$ Option 4) $\sqrt{x}-\left ( x+1 \right )tan^{-1}\sqrt{x}+C$

G gaurav

As we learnt

Rule for integration by parts -

 Take Ist function $(u)$ as according I L A T E

- wherein

 Where , I : Inverse L : Logarithmic A : Algebraic  T : Trignometric E : Exponential

$I = \int {{{\tan }^{ - 1}}\sqrt x \,\,dx = x.{{\tan }^{ - 1}}\sqrt x } - \int {\frac{1}{{1 + x}}.\frac{2}{{2\sqrt x }}} .x\,\,dx = x.{\tan ^{ - 1}}\sqrt x - \frac{1}{2}\int {\frac{{\sqrt x }}{{1 + x}}} \,dx$

Now putting $\sqrt x = t$ i.e., $\frac{1}{{2\sqrt x }}\,dx = dt$  Þ  $dx = 2tdt$, we get

$I = x.{\tan ^{ - 1}}\sqrt x - \frac{1}{2}\int {\frac{t}{{1 + {t^2}}}2t} \,\,dt = x.{\tan ^{ - 1}}\sqrt x - \int {\frac{{{t^2} + 1 - 1}}{{1 + {t^2}}}} \,\,dt$

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

Option 1)

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

Option 2)

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

Option 3)

$\sqrt{x}-xtan^{-1}\sqrt{x}+C$

Option 4)

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

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