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\int e^{x}\,\frac{x-1}{(x+1)^{3}} dx=

  • Option 1)

    -\frac{e^{x}}{x+1}+c

  • Option 2)

    \frac{e^{x}}{x+1}+c

  • Option 3)

    \frac{e^{x}}{(x+1)^{2}}+c

  • Option 4)

    -\frac{e^{x}}{(x+1)^{2}}+c

 

Answers (1)

best_answer

As 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

 

 

=\int e^{x} \: \frac{x-1}{(x+1)^{3}} \: dx

=\int e^{x} \left [ \frac{x+1-2}{(x+1)^{3}} \right ]dx

=\int e^{x} \left [ \frac{1}{(x+1)^{2}}-\frac{2}{(x+1)^{3}} \right ]dx

=e^{x} \cdot \frac{1}{(x+1)^{2}}+c          \\ \left [ \because \: \: \int e^{x}(f(x)+f^{1}(x)) dx \right ] \\ \\ =e^{x}f(x)+c

 


Option 1)

-\frac{e^{x}}{x+1}+c

This option is incorrect

Option 2)

\frac{e^{x}}{x+1}+c

This option is incorrect

Option 3)

\frac{e^{x}}{(x+1)^{2}}+c

This option is correct

Option 4)

-\frac{e^{x}}{(x+1)^{2}}+c

This option is incorrect

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divya.saini

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