Question

Let $I=\int \left(x^2+1\right){\left((x+1)e^x\right)}^{2}dx=A{f}^2(x)+C$, where $C$ is a constant of integration and $f(-1)=\left(\frac{2}{e}\right)$, then value of $2A+f(0)$ 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 $\left(x^2+1\right)e^x=t\Rightarrow (x+1{)}^2{e}^{x}dx=dt$
$\Rightarrow \int tdt=\frac{t^2}{2}+c\Rightarrow I=\frac{{\left(\left(x^2+1\right)e^x\right)}^2}{2}+c$
$\Rightarrow f(x)=\left(x^2+1\right)e^x\text{ and }A=\frac{1}{2}$
$\Rightarrow f(0)=1$
$\Rightarrow 2A+f(0)=2$