Question

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

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

Answer 1

Answer: (D)

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

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

$\Rightarrow \int e^{x}dx+\int 2e^x(-\frac{1}{x+1}+\frac{1}{(x+1{)}^2})dx$ ( doing the partial fraction of $\frac{-x}{(x+1{)}^2}$)

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