Question

$\int e^x{\left(\frac{1-x}{1+x^2}\right)}^{2}dx$ is equal to

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

Answer 1

Answer: (A)

We have, $I=\int e^x{\left(\frac{1-x}{1+x^2}\right)}^{2}dx$

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

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

Above integral is of the type $\int e^x\left(f\left(x\right)+f^{\prime }\left(x\right)\right)dx$

$\therefore$ Solution is $e^{x}f\left(x\right)+C$

$\Rightarrow I=e^x\left(\frac{1}{1+x^2}\right)+C$