Question

Let f(x) be the function part of the integral part, the find f(0)

$\int \frac{e^x(x^3+x+1)}{{(x^2+1)}^{3/2}}dx$

Answer 1

$I=\int \frac{e^x(x^3+x+1)}{{(x^2+1)}^{3/2}}dx$
$=\int e^x[\frac{x}{\sqrt{x^2+1}}+\frac{1}{{(x^2+1)}^{3/2}}]$
It can be easily seen that
$\frac{d}{dx}\left\{\frac{x}{\sqrt{(x^2+1)}}\right\}=\frac{1}{{(x^2+1)}^{3/2}}$
$\therefore I=\int e^x[f\left(x\right)+f^{\prime }\left(x\right)]dx=e^{x}f\left(x\right)$
$\therefore I=e^x.\frac{x}{\sqrt{(x^2+1)}}$