Integral (square root function multiplied by exponential function) did i do it right? I'm trying to determine $\int x^3\sqrt{x^2 +1}\, dx$
I said that $u(x) = x^2 + 1$
and then that $dx = 2x\,dx$
so I rewrote the integral as 
$$\int x^3\sqrt{x^2 +1}\,2x\,dx$$
which is also 
$$\int2x^4\sqrt{x^2 +1} \,dx$$
and then it is easy to integrate, is that all legal to do?

\begin{align}
&\int x^{3}\,\sqrt{x^{2} + 1\,}\,{\rm d}x
=
\int x^{2}\,{\rm d}\left[{1 \over 3}\,\left(x^{2} + 1\right)^{3/2}\right]
\\[3mm]&=
x^{2}\,{1 \over 3}\left(x^{2} + 1\right)^{3/2}
-
\int{1 \over 3}\left(x^{2} + 1\right)^{3/2}
\,{\rm d}\left(x^{2} + 1\right)
\\[3mm]&=
{1 \over 3}\,x^{2}\left(x^{2} + 1\right)^{3/2}
-
{1 \over 3}\,{\left(x^{2} + 1\right)^{5/2} \over 5/2}
=
\left(x^{2} + 1\right)^{3/2}\left[%
{1 \over 3}\,x^{2}
-
{2 \over 15}\left(x^{2} + 1\right)
\right]
\\[3mm]&=
\left(x^{2} + 1\right)^{3/2}\,{3x^{2} - 2 \over 15}
\end{align}
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
\int x^{3}\,\sqrt{x^{2} + 1\,}\,{\rm d}x
\color{#000000}{\ =\ }
{1 \over 15}\left(3x^{2} - 2\right)\left(x^{2} + 1\right)^{3/2}\
+\
\color{#000000}{\mbox{constant}}
\quad}
\\ \\ \hline
\end{array}
$$
