Question

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

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

Answer 1

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

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

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

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

$=\int e^x[f\left(x\right)+f^{\prime }\left(x\right)]dx,$

Where $f\left(x\right)=\frac{x-1}{x+1}$ and $f^{\prime }\left(x\right)=\frac{2}{{(x+1)}^2}$

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