$$\int x\,dx^2$$
I'm having trouble comprehending this question. I can perform substitution, but when I do I come out wrong. Here's how I've happened to handle this:
$\int x\,dx^2$
$u=x^2$
$du=2x\,dx$
$dx=\frac {du}{2x}$
$\frac12 \int x\frac{du}{2x}$
$\frac 12\int\frac12\,du$
$\frac u4$
$\frac {x^2}{4}$
It says that it's wrong. Me showing my work was just a desperate attempt at solving it. I'd really like a guide through the problem please.
Also, apparently the answer is $\frac {2x^3}{3}$
$\endgroup$ 65 Answers
$\begingroup$If $u = x^2$, then modulo a sign, $x = \sqrt{u}$. Hence the integral is equal to
$$\int \sqrt{u} \ du = \frac{2}{3}u^{3/2} \ (+ C) = \frac{2}{3}x^3 \ (+ C)$$
$\endgroup$ 0 $\begingroup$An alternative approach: $d(x^2) = 2x\,dx$ so you have
$$\int x\,d(x^2) = \int 2x^2 \,dx = \frac{2}{3}x^3+C.$$
$\endgroup$ $\begingroup$Everything was going fine until $$\mathrm{d}x=\frac{\mathrm{d}u}{2x}$$. Instead, letting $u=x^2\implies x=\sqrt{u}$ $$\int\!x\,\mathrm{d}(x^2)=\int\!\sqrt{u}\,\mathrm{d}u$$ $$=\frac{2}{3}\sqrt{u}^3+c$$ $$=\frac{2}{3}x^3+c$$
$\endgroup$ 1 $\begingroup$you can solve this integration by another method. As we are familiar with integrating with dx instead of $dx^2$, so it will be easier if we translate it to an integral with dx.
Let $y = \int x\,dx^2$ so $\frac{dy}{dx^2}=x$ then we divide by $dx$ to get $\frac{\frac{dy}{dx}}{\frac{dx^2}{dx}} = x$. $\frac{dx^2}{dx}=2x$, that mean $\frac{y'}{2x}=x$ so $\frac{dy}{dx}=2x^2$ so $y = \frac{2x^3}{3}+k$
$\endgroup$ $\begingroup$It is very simple:
$$\int xdx^2 = \int \sqrt{x^2}dx^2=\frac{2}{3}x^3 + c$$
$\endgroup$
