Question

The value of $\int e^x.\frac{x^2+1}{{(x+1)}^2}dx$ is

• A. $e^x\left(\frac{x-1}{x+1}\right)+C$
• B. $e^x\left(\frac{x+1}{x-1}\right)+C$
• C. $e^x.x+C$
• D. None of these

Answer 1

The correct option is A $e^x\left(\frac{x-1}{x+1}\right)+C$

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

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

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

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

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

$=e^x-I_1+e$

Let $I_1=\int e^x\cdot \frac{2x}{(x+1{)}^2}dx$

Here $\frac{2x}{(x+1{)}^2}=\frac{A}{(x+1)}+\frac{B}{(x+1{)}^2}$

$2x=A(x+1)+B$

If $x=-1$,

$2(-1)=B$

$\Rightarrow B=-2$

If $x=1$,

$2=2A+(-2)$

$2A=2+2$

$A=\frac{4}{2}=2$

$\therefore \frac{2x}{(x+1{)}^2}=\frac{2}{x+1}-\frac{2}{(x+1{)}^2}$

$\therefore I_1=\int e^x\{\frac{2}{x+1}-\frac{2}{(x+1{)}^2}\}dx$

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

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

Which is of the form $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx$

$=2e^x\cdot \left(\frac{1}{x+1}\right)$ $[\because \int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)+c]$

$=\frac{2e^x}{x+1}$

$\therefore I=e^x-\frac{2e^x}{x+1}+c$

$=e^x(1-\frac{2}{x+1})+c$

$=e^x\left\{\frac{x+1-2}{x+1}\right\}+c$

$\therefore I=e^x\left(\frac{x-1}{x+1}\right)+c$.