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?
 A: 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.
A: $u$-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
$$\begin{align*}
f'(x) &= \sqrt{2x}\\
&= \sqrt2 \cdot \sqrt x\\
f(x) &= \sqrt2\cdot\frac{2}3 x^{3/2}+C\\
f(g(x)) &= \sqrt2\cdot\frac23\left[\frac12(1+x^2)\right]^{\frac32}+C\\
&= \frac13(1+x^2)^{3/2}+C
\end{align*}$$
A: Knowing that a derivative of powers reduces that power by one leads one to consider $$\frac{d}{dx} \, (1 + x^2)^{3/2} = 3 \, x \, \sqrt{1+x^2}.$$ 
From this it is easily seen that
$$\int x \, \sqrt{1 + x^2} \, dx = \frac{1}{3} \, (1 + x^2)^{3/2} + c_{0}.$$
A: Alternatively:
$$\int{x\sqrt{1+x^2} \,\mathrm{d}x} = \frac12\int{\sqrt{1+x^2} \,\mathrm{d}x^2}=\\
\frac12\int \sqrt{1+t} \, \ \mathrm dt=\frac12\cdot \frac23\cdot (1+t)^{3/2}+C=\frac13(1+x^2)^{3/2}+C.$$
A: Whenever you are finding the integral of a function which consists of  f(x) & f'(x) of some function ( Here f(x)=x^2 & f'(x)=2x)
Always go by substituting f(x) to some other variable
Put t^2 = 1+x^2 (I'm taking like this because square root of t^2 will result t which makes my integration easier)
Then 2t dt = 2x dx 
 ==> t dt = x dx 
Substitute these values in the integration..
∫x√(1+x^2) dx = ∫√t^2 t dt
                          =∫t×t dt
                          =∫t^2 dt
                          =(t^3)/3
Finally substituting t = √(1+x^2) gives the anti derivative ((1+x^2)^(3/2))/3
