<s>[INST] 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?
 [/INST] Alternatively:
$$\int{x\sqrt{1+x^2} \,\mathrm{d}x} = \frac12\int{\sqrt{1+x^2} \,\mathrm{d}x^2}=\\
\frac12\int \sqrt{1+t} \, \ \mathrm dt=\frac12\cdot \frac23\cdot (1+t)^{3/2}+C=\frac13(1+x^2)^{3/2}+C.$$
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