<s>[INST] Integration-by-part Here is the given question 
$$\int{\frac{x^3}{\sqrt{1+x^2}}dx}$$
I solved using integration by part as follow:
$$
\int\frac{x}{\sqrt{1+x^2}}x^2\,dx = x^2\int\frac{x}{\sqrt{1+x^2}}\,dx - \int\biggl(\frac{dx^2}{dx}\int\frac{x}{\sqrt{1+x^2}}dx\biggr)dx\tag{i}\label{i}
$$
Solving for $\int \frac{x}{\sqrt{1+x^2}}$ and taking $1+x^2 = t \rightarrow 2xdx = dt \rightarrow xdx = \frac{dt}{2}$.
\begin{align}
&\int \frac{xdx}{\sqrt{1+x^2}} = \int\frac{dt}{2\sqrt{t}}\\
&\int\frac{dt}{2\sqrt{t}} = \frac{1}{2}\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} \rightarrow \sqrt{t} = \sqrt{1+x^2}
\end{align} 
Pluggin in equation \eqref{i}
$$\int{\frac{x}{\sqrt{1+x^2}}x^2dx} = x^2\sqrt{1+x^2} - 2\int{x\sqrt{1+x^2}dx}\tag{ii}\label{ii}$$
Now solving for $\int{x\sqrt{1+x^2}dx}$.
Assuming $1+x^2 = t \rightarrow xdx = \frac{dt}{2}$
$$\frac{1}{2}  \int\sqrt{t}dt \rightarrow \frac{\sqrt[3]{t}}{3} \text{  or   }
\int{x\sqrt{1+x^2}dx} =  \frac{\sqrt[3]{1+x^2}}{3}$$
Pluggin in equation  
$$\int\frac{x}{\sqrt{1+x^2}}x^2dx = x^2\sqrt{1+x^2} - 2 \frac{\sqrt[3]{1+x^2}}{3} + C$$
But given answer is:
$$\frac{1}{3}(1+x^2)^{3/2}-(1+x^2)^{1/2}+C$$
What is wrong with my solution?
 [/INST] It should be
$$\int{\frac{x}{\sqrt{1+x^2}}*x^2dx} = x^2\sqrt{1+x^2} - 2\cdot \frac{(1+x^2)^{3/2}}{3} + C$$
Use 
$$x^2\sqrt{1+x^2}=(x^2+1-1)\sqrt{1+x^2}=(1+x^2)^{3/2}-(1+x^2)^{1/2}$$ to match with the given answer
</s>