#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 38 Maths Textbook Solution.

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

Given:

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

Hint:

Using integration by parts.

Explanation:

Let $I=\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x$                                                         $\left[\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right]$

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

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

\begin{aligned} &=\int \frac{e^{x}}{\left(1+x^{2}\right)} d x-\int \frac{e^{x} 2 x}{\left(1+x^{2}\right)^{2}} d x \\ &=\int e^{x} \cdot \frac{1}{1+x^{2}} d x-\int e^{x} \cdot \frac{2 x}{\left(1+x^{2}\right)^{2}} d x \end{aligned}

$=\frac{1}{1+x^{2}} \cdot e^{x}-\int e^{x}\left(\frac{\left(1+x^{2}\right)(0)-(1)(2 x)}{\left(1+x^{2}\right)^{2}}\right) d x-\int e^{x} \frac{2 x}{\left(1+x^{2}\right)^{2}} d x\left\{\int u . v d x=u \int v d x-\int \frac{d u}{d x} \int v d x d x\right\}$

\begin{aligned} &=\frac{e^{x}}{1+x^{2}}+\int \frac{2 x e^{x}}{\left(1+x^{2}\right)^{2}} d x-\int \frac{2 x e^{x}}{\left(1+x^{2}\right)^{2}} d x \\ &=\frac{e^{x}}{1+x^{2}}+C \end{aligned}