Question

$\int \frac{\left(x^{2}+1\right) e^{x}}{(x+1)^{2}} d x=f(x) e^{x}+C$, 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)

$\int\left(\frac{x^{2}+1}{(x+1)^{2}}\right) e^{x} \cdot d x$
$=\int\left(\frac{x^{2}-1+2}{(x+1)^{2}}\right) e^{x} d x$
$=\int\left(\frac{x-1}{x+1}+\frac{2}{(x+1)^{2}}\right) e^{x} d x$
$=\int\left(f(x)+f^{\prime}(x)\right) e^{x} d x$
$=f(x) e^{x}+c$
Where $f(x)=\frac{x-1}{x+1}$
$f^{\prime}(x)=\frac{2}{(x+1)^{2}}$
$f^{\prime \prime}(x)=\frac{-4}{(x+1)^{3}}$
$=\frac{12}{(x+1)^{4}}$
$f^{\prime \prime}(1)=\frac{12}{16}$
$=\frac{3}{4}$