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\int e^{x}\left ( \tan^{-1}x+\frac{2x}{\left ( 1+x^{2} \right )^{2}} \right )dx=

  • Option 1)

    e^{x}\left ( \tan^{-1}x-\frac{1}{1+x^{2}} \right )+C

  • Option 2)

    e^{x}\left ( \tan^{-1}x+\frac{1}{1+x^{2}} \right )+C

  • Option 3)

    e^{x}\left ( \cot^{-1}x-\frac{1}{1+x^{2}} \right )+C

  • Option 4)

    e^{x}\left ( \tan^{-1}x+\frac{2}{1+x^{2}} \right )+C

 

Answers (1)

best_answer

As we learnt

Result for integration by parts -

\int e^{x}(f(x)+f'(x))dx=e^{x}f(x)+c

- wherein

It works on integration by parts of  \int e^{x}f\left ( x \right )dx

 

  

 

Result for integration by parts -

\int e^{x}(f(x)+f'(x))dx=e^{x}f(x)+c

- wherein

It works on integration by parts of  \int e^{x}f\left ( x \right )dx

 

\int {{e^x}\left( {{{\tan }^{ - 1}}x + \frac{1}{{1 + {x^2}}} - \frac{1}{{1 + {x^2}}} + \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}} \right)dx}= {e^x}\left( {{{\tan }^{ - 1}} - \frac{1}{{1 + {x^2}}}} \right) + C


Option 1)

e^{x}\left ( \tan^{-1}x-\frac{1}{1+x^{2}} \right )+C

Option 2)

e^{x}\left ( \tan^{-1}x+\frac{1}{1+x^{2}} \right )+C

Option 3)

e^{x}\left ( \cot^{-1}x-\frac{1}{1+x^{2}} \right )+C

Option 4)

e^{x}\left ( \tan^{-1}x+\frac{2}{1+x^{2}} \right )+C

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