Integrating a product, one factor a derivative I'm solving a differential equation and have to integrate this term:
$\int \frac{dx(t)}{dt} x(t)^2 dt$
Partial integration gave me $0$ as result, so I gave it a try on wolframalpha. This came up with a solution that is analog to results I've seen for similar equations (The derivative just seems to equal to $1$ and we use a usual integration).
Wolframalphas solution: $\int \frac{dx(t)}{dt} x(t)^2 dt = \frac{x(t)^3}{3} + c$
My problem is, I can't follow the steps wolframalpha shows.
Why can you substitute like wolframalpha does? What are the rules used to do that?
 A: If integration by parts gave you $0$, it's probably because you integrated wrong.
The simplest way to do this problem is by substitution: let $u=x(t)$. Then $du = x'(t)\,dt = \frac{dx}{dt}\,dt$. 
So we can do a simple substitution:
$$\int\frac{dx(t)}{dt}(x(t))^2\,dt = \int u^2\,du = \frac{1}{3}u^3+C = \frac{1}{3}(x(t))^3 + C.$$
As for integration by parts, if we set $u=(x(t))^2$, $dv = \frac{dx(t)}{dt}\,dt$, then we can take $v=x(t)$, $du = 2x(t)x'(t)\,dt$, so we would get
$$\int\frac{dx(t)}{dt}(x(t))^2\,dx = (x(t))^3 - \int 2(x(t))^2\frac{dx(t)}{dt}\,dt.$$
Note that the integral on the right is the same as the integral on the left, but multiplied by $-2$; if we move it to the left hand side, we obtain
$$\int\frac{dx(t)}{dt} (x(t))^2\,dt + 2\int\frac{dx(t)}{dt}(x(t))^2|,dt = (x(t))^3+C.$$
Now adding the two integrals and dividing by three we get
$$\begin{align*}
3\int\frac{dx(t)}{dt}(x(t))^2\,dt &= (x(t))^3 + C\\
\int\frac{dx(t)}{dt}(x(t))^2\,dt &= \frac{1}{3}(x(t))^3 + c.
\end{align*}$$
A: I’m going to assume that $\dfrac{x(t)}{dt}$ is a typo for $\dfrac{d(x(t))}{dt}$.
Look at a specific example, say with $x(t)=\sin t$. Then $$\int\frac{d(x(t))}{dt}x(t)^2dt=\int\cos t\,\sin^2tdt\;,$$ a problem that you would most likely solve by making the substitution $u=\sin t$, $du=\cos tdt$, and integrating $$\int u^2 du\;.$$
But you don’t have to know what $x(t)$ is to make this substitution. If you let $u=x(t)$, then $$du=\frac{d(x(t))}{dt}dt\;,$$ and $$\int\frac{d(x(t))}{dt}x(t)^2dt=\int u^2 du=\frac{u^3}3+C=\frac{x(t)^3}3+C\;.$$
A: Also, since $\int \frac{dx(t)}{dt} x(t)^2 dt = \int \frac{d f(t)}{dt} dt$, where $f(t) = \frac{1}{3} x(t)^3$, we can use the Second Fundamental Theorem of Calculus to conclude that $\int \frac{dx(t)}{dt} x(t)^2 dt = f(t) + C = \frac{1}{3} x(t)^3 + C$. No substitution needed.
