Question

The value of the integral $\int e^x{\left(\frac{1-x}{1+x^2}\right)}^{{}^2}$ dx is

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

Answer 1

Answer: (C)

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