Question

Let $I=\int\left(x^2+1\right)\left((x+1) e^x\right)^2 d x=Af^2(x)+C$, where $C$ is a constant of integration and $f (-1)=\left(\frac{2}{ e }\right)$, then value of $2 A + f (0)$ equals

• A. 3
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (B)

$\int\left( x ^2+1\right) e ^{ x } \cdot( x +1)^2 e ^{ x } dx$
Let $\left( x ^2+1\right) e ^{ x }= t \Rightarrow( x +1)^2 e ^{ x } dx = dt$
$\Rightarrow \int tdt =\frac{ t ^2}{2}+ c \Rightarrow I =\frac{\left(\left( x ^2+1\right) e ^{ x }\right)^2}{2}+ c$
$\Rightarrow f ( x )=\left( x ^2+1\right) e ^{ x } \text { and } A =\frac{1}{2} $
$\Rightarrow f (0)=1 $
$\Rightarrow 2 A + f (0)=2$