Question

$\int e^x\frac{(x^2+1)}{{(x+1)}^2}dx$ is equal to

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

Answer 1

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

$I=\int e^x\frac{(x^2+1)}{{(x+1)}^2}dx=\int \frac{e^x(x^2-1+2)}{{(x+1)}^2}dx=\int e^x[\frac{x-1}{x+1}+\frac{2}{{(x+1)}^2}]dx=\int e^x[f\left(x\right)+f^{\prime }\left(x\right)]dx,\text{where}f\left(x\right)=\frac{x-1}{x+1}\text{and}{f}^{\prime }\left(x\right)=\frac{2}{{(x+1)}^2}\{\because \int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+C\}\Rightarrow I=e^x\left(\frac{x-1}{x+1}\right)+C$