Question

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

• A. $e^x\cdot \frac{x+1}{x-1}.$
• B. $-e^x\cdot \frac{x-1}{x+1}.$
• C. $e^x\cdot \frac{x-1}{x+1}.$
• D. $\frac{1}{2}{e}^x\cdot \frac{x-1}{x+1}.$

Answer 1

Answer: (C)

$e^x\cdot \frac{x-1}{x+1}.$

$=\frac{x^2+1}{{(x+1)}^2}=\frac{x^2-1}{{(x+1)}^2}=\frac{x-1}{x+1}+\frac{2}{{(x+1)}^2}=f\left(x\right)+f^{\prime }\left(x\right)$
$\therefore I=\int e^x[f\left(x\right)+f^{\prime }\left(x\right)]dx=e^{x}f\left(x\right)=e^x.\frac{x-1}{x+1}.$
Alternative Method:
$I=\frac{e^x(x^2+1)}{{(x+1)}^2}.$ Integrate by parts
$I=e^x(x^2+1)[-\frac{1}{x+1}]+\int \frac{1}{x+1}.e^x[x^2+1+2x]dx$
$I=-e^x.\frac{x^2+1}{x+1}+\int e^x(x+1)dx=-e^x.\frac{x^2+1}{x+1}+e^x.x.$
$e^x.\left[\frac{-x^2-1+x^2+x}{x+1}\right]=e^x\cdot \frac{x-1}{x+1}.$