<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] $u$-substitution is more standard, but I can follow your method to see how it works out.
$$\begin{align*}
\frac d{dx}f(g(x)) &= g'(x) \cdot f'(g(x))\\
&= x\sqrt{1+x^2}
\end{align*}$$
Try $g'(x) = x$, then
$$g(x) = \frac12x^2+C_1$$
Try $C_1 = \frac12$ to get $g(x) = \frac12(1+x^2)$. Then
$$\begin{align*}
f'(x) &= \sqrt{2x}\\
&= \sqrt2 \cdot \sqrt x\\
f(x) &= \sqrt2\cdot\frac{2}3 x^{3/2}+C\\
f(g(x)) &= \sqrt2\cdot\frac23\left[\frac12(1+x^2)\right]^{\frac32}+C\\
&= \frac13(1+x^2)^{3/2}+C
\end{align*}$$
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