Question

$$\begin{gathered} \int e^x{\left(\frac{1-x}{1+x^2}\right)}^{2}dx\mathrm{is}\mathrm{equal}\mathrm{to} \\ \left(\mathrm{a}\right)\frac{e^x}{1+x}+C\left(\mathrm{b}\right)-\frac{e^x}{1+x^2}+C\left(\mathrm{c}\right)\frac{e^x}{{(1+x}^{}} \end{gathered}$$²)2+C (d) -ex(1+x²)2+C

Answer 1

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

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

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

Let $f\left(x\right)=\frac{1}{1+x^2}$

$\Rightarrow f\text{'}\left(x\right)=\frac{-2x}{{\left(1+x^2\right)}^2}$

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

Hence, the correct answer is option (c).