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  • Confused! kindly explain, - Integral Calculus - JEE Main-8

The integral \int e^{\tan^{-1}x}\left ( \frac{1+x+x^{2}}{1+x^{2}} \right )dx is equal to

  • Option 1)

    \frac{e^{\tan^{-1}x}}{1+x^{2}} +C

  • Option 2)

    xe^{\tan^{-1}x}+C

  • Option 3)

    \frac{xe^{\tan^{-1}x}}{1+x^{2}} +C

  • Option 4)

    none

 

Answers (1)

best_answer

As we learnt

Special type of indefinite integration -

Integration of the form : g(sin^{-1}f(x))

Put f(x)=t 

so f'(x)dx=dt

Now Put \Theta =sin^{-1}t

such that t=sin\Theta

-

 

 Putting tan-1x = u, we have   \frac{{{\rm{dx}}}}{{{\rm{1}} + {{\rm{x}}^{\rm{2}}}}}=du

\int {{{\rm{e}}^{{\rm{ta}}{{\rm{n}}^{ - {\rm{1}}}}{\rm{x}}}}\left( {\frac{{{\rm{1}} + {\rm{x}} + {{\rm{x}}^{\rm{2}}}}}{{{\rm{1}} + {{\rm{x}}^{\rm{2}}}}}} \right)}dx=\int {{e^u}\left( {1 + \tan u + {{\tan }^2}u} \right)du} $

$\int {{e^u}} $(sec2u + tan u)du = tan u eu + C = x ${e^{{{\tan }^{ - 1}}x}}$ + C

 


Option 1)

\frac{e^{\tan^{-1}x}}{1+x^{2}} +C

Option 2)

xe^{\tan^{-1}x}+C

Option 3)

\frac{xe^{\tan^{-1}x}}{1+x^{2}} +C

Option 4)

none

Posted by

Aadil

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