Question

Evaluate $\int e^x{\left(\frac{1-x}{1+x^2}\right)}^{2}dx$
(where $C$ is constant of integration)

• A. $\frac{e^x}{1+x^2}+C$
• B. $\frac{e^x}{1-x^2}+C$
• C. $-\frac{e^x}{1+x^2}+C$
• D. $-\frac{e^x}{1-x^2}+C$

Answer 1

The correct option is A $\frac{e^x}{1+x^2}+C$

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