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\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

 

Answers (1)

best_answer

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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gaurav

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