Question

The value of the integral $\underset{-1}{\overset{1}{\int }}\{\frac{x^{2015}}{e^{|x|}(x^2+\cos x)}+\frac{1}{e^{|x|}}\}dx$ is equal to

• A. $0$
• B. $1-e^{-1}$
• C. $2e^{-1}$
• D. $2(1-e^{-1})$

Answer 1

The correct option is D $2(1-e^{-1})$

$\underset{-1}{\overset{1}{\int }}\{\frac{x^{2015}}{e^{|x|}(x^2+\cos x)}+\frac{1}{e^{|x|}}\}dx$
First function is odd and integral is from $-1$ to $1$, so its is $0.$
$=0+2\underset{0}{\overset{1}{\int }}\frac{1}{e^x}dx$
$=-2e^{-x}{|}_0^1$
$=-2(e^{-1}-1)$
$=2(1-e^{-1})$