Antiderivative of $x\sqrt{1+x^2}$ I am attempting this problem given to me, but the answer key does not explain the answer.  The question asks me to find the antiderivative of $x\sqrt{1+x^2}$.
My attempt:
$\frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$
$g'(x)$ must be $x$, therefore $g(x)=\frac{1}{2}x^2$
$f'(x)$ must equal $\sqrt{1+x^2}$, therefore $f(x)=\frac{2}{3}(1+x^2)^{\frac{3}{2}}$
But then I saw that if $g(x)=\frac{1}{3}x^2$, then $g'(x)=\frac{2}{3}x$,  and $f(x)=(1+x^2)^{\frac{3}{2}}$, then $\frac{d}{dx}f(g(x))'=g'(x)f'(g(x))=(2/3)x(3/2)\sqrt{1+x^2}=x\sqrt{1+x^2}$.
But I saw this by chance; what would be a more general and effective way to do this?  Should I check $this$, and then write $that$, and put together $those$, etc.? Or is the way I just 'saw' the way the numbers should fall together how one would normally approach this?

You're effectively doing a $u$-substitution to compute 
$$\int{x\sqrt{1+x^2} \,\mathrm{d}x}$$
 with $u=1+x^2$ and hence $du=2xdx$. This turn the earlier integral into 
$$\int{\frac{1}{2}u^{1/2}\,\mathrm{d}u}$$
We can easily integrate this to get
$$\frac{2}{3}\frac{1}{2}u^{3/2}+C = \frac{1}{3}(1+x^2)^{3/2}+C$$
$u$-substitution can be thought of as being the integral-version of chain rule, much in the same way that integration by-parts is the integral-version of product rule.
