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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 of these

 

Answers (1)

best_answer

As we learnt

Result for integration by parts -

e^{f(x)}\left [ g(x)f'(x)+g'(x) \right ]dx=e^{f(x)}g(x)+c

- wherein

or ex :

\int e^{tan^{-1}x}[\frac{x^{n}+1}{x^{2}+1}+nx^{n-1}]dx=e^{tan^{-1}x}(x^{n}+1)+c

 

 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}} \left( {se{c^2}u + {\rm{ }}tanu} \right)du = {\rm{ }}tanu{e^u} + {\rm{ }}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 of these

Posted by

gaurav

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