Question

If $\int \frac{\left(x^2+1\right)e^x}{(x+1{)}^2}dx=f(x)e^x+C$, where $C$ is a constant, then $\frac{d^{3}f}{d{x}^3}$ at $x=1$ is equal to

• A. $-\frac{3}{4}$
• B. $\frac{3}{4}$
• C. $-\frac{3}{2}$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

$I=\int \frac{e^x\left(x^2+1\right)}{(x+1{)}^2}dx=f(x)e^x+c$
$=\int \frac{e^x\left(x^2-1+1+1\right)}{(x+1{)}^2}dx$
$=\int e^x\left[\frac{x-1}{x+1}+\frac{2}{(x+1{)}^2}\right]dx$
$=e^x\left(\frac{x-1}{x+1}\right)+c$
$\therefore f(x)=\frac{x-1}{x+1}$
$f(x)=1-\frac{2}{x+1}$
$f^{\prime }(x)=2{\left(\frac{1}{x+1}\right)}^2$
$f^{\prime \prime }(x)=-4{\left(\frac{1}{x+1}\right)}^3$
$f^{\prime \prime \prime }(x)=\frac{12}{(x+1{)}^4}$
$\text{ for }x=1$
$f^{\prime \prime \prime }(1)=\frac{12}{2^4}=\frac{12}{16}=\frac{3}{4}$