Question

$$\int e^x\frac{{(1-x)}^2}{{(1+x^2)}^2}\text{ dx }$$

Answer 1

$$\text{We have},$$

$$I=\int \frac{e^x{(1-x)}^2}{{(1+x^2)}^2}\text{ dx }$$

$$=\int \frac{e^x(1+x^2-2x)}{{(1+x^2)}^2}\text{ dx }$$

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

$$=\int e^x(\frac{1}{1+x^2}-\frac{2x}{{(1+x^2)}^2})\text{ dx }$$

$$=\frac{e^x}{1+x^2}+C.......................[\because \int e^{x}f(x)+f^{\prime }(x)\text{ dx }=e^{x}f(x)+C$$${\displaystyle \text{ Where},ƒ \left(x\right\}= \frac{1}{1+2}\Rightarrow {ƒ}^{\prime } \left(x\right) =- \frac{2x}{{(1+x^2)}^2}]}$

 

 

Answer 2

$$\text{We have},$$

$$I=\int \frac{e^x{(1-x)}^2}{{(1+x^2)}^2}dx$$

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

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

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

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