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 $\int e^{x}\left( f\left(x\right) +f'\left(x\right)\right) dx$

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

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