Question

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

Answer 1

\[\text{We have}, \]

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

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

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

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

\[ = \frac{e^x}{1 + x^2} + C.......................[ ∵  \int e^x { f  ( x ) + f' ( x ) }  \text{ dx }= e^x f( x ) + C\]`   \text{  Where} ,ƒ  (x } =   1/{1+2} ⇒ ƒ ^'  (x )  = -  {2x}/ (1+ x^2)^2 ]`

 

 

Answer 2

\[\text{We have}, \]

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

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

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

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

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