Answer
$\frac{1}{5}{(1+x^2)}^{\frac{5}{2}}-\frac{1}{3}{(1+x^2)}^{\frac{3}{2}}+C$
This answer is reasonable because $f(x)=F'(x)$
See image below
Work Step by Step
$\int x^3\sqrt{1+x^2}$
Substitute $\displaystyle \left[\begin{array}{ll}
a=1+x^2, & a-1=x^2\\
& \\
\frac{da}{dx}=2x & \frac{da}{2x}=dx
\end{array}\right]$
$I=\int x(a-1)\sqrt{a}\frac{da}{2x}\\
I=\int\frac{1}{2}(a-1)a^{\frac{1}{2}}da\\
I=\frac{1}{2}\int{(a^{\frac{3}{2}}-a^{\frac{1}{2}})da}\\
I=\frac{1}{2}{(\frac{2}{5}a^{\frac{5}{2}}-\frac{2}{3}a^{\frac{3}{2}})}+C\\
I=\frac{1}{5}a^{\frac{5}{2}}-\frac{1}{3}a^{\frac{3}{2}}+C\\
I=\frac{1}{5}{(1+x^2)}^{\frac{5}{2}}-\frac{1}{3}{(1+x^2)}^{\frac{3}{2}}+C$
