Find integral when $dx$ is in the numerator

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Can someone please walk me through the steps to find the following integral? I'm not sure what to do when $dx$ is at the top.

$$ \int \frac{ x^{2}dx }{ (x^{3} + 5)^{2}} $$

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The $dx$ inside an integral expression can appear in various places, even if when integrals are introduced, it is (usually) always written at the end. An informal but intuitive (but probably dangerous!) way is to think of the expression as a multiplication of the integrand by the differential $dx$.

So

$$ \int \frac{ x^{2}dx }{ (x^{3} + 5)^{2}} = \int \frac{ x^{2} }{ (x^{3} + 5)^{2}} dx.$$

In fact, many people, at least many physicists, often write e.g. $$ \int_a^b dx\left\{ \frac{ x^{2} }{ (x^{3} + 5)^{2}} \right\},$$ so the reader (one who is familiar with this notation) will immediately see what is the variable of integration.

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What do you mean? It is its "usual place", $\int_D f(x)dx$...

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Put $u = x^3 + 5$ and then we will have $du = 3x^2dx$. Thus we have $x^2dx = du/3$. Thus, we will get the integral $$\frac{1}{3}\int \frac{du}{u^2} $$ which you can then evaluate.

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Having $dx$ in the numerator is the same as having it appear as if it is multiplying the integrand:

$$ \int \frac{ x^{2}dx }{ (x^{3} + 5)^{2}} = \int \frac{ x^{2} }{ (x^{3} + 5)^{2}}dx$$

So just start with something like $u = x^3 + 5$ and $du = 3x^2 dx$, and go to town.