Question

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

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