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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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It is different notation for $\int \dfrac{ x^{2}}{ (x^{3} + 5)^{2}}dx$, no more than that. – Pedro Tamaroff Feb 8 '14 at 22:43
An fyi, seeing as this is now a "hot question": The "top" of a fraction is called the numerator, not enumerator. (An enumerator is something that labels/counts things.) – apnorton Feb 9 '14 at 4:17
up vote 10 down vote accepted

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 $dx$.


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

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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Well, JiK used the the cautionary formulation "can be thought of as" not "is" – Hagen von Eitzen Feb 8 '14 at 22:48
@JonathanY. That's true but unfortunately I don't know a better (while still simple) way to explain why $dx$ can appear almost anywhere in the expression. – JiK Feb 8 '14 at 22:49
I rewrote the first paragraph. Thanks for comments! – JiK Feb 8 '14 at 23:07
FWIW in physics we almost always think of integrals as limits of Riemann sums, and in a Riemann sum the width $\Delta x$ is just a factor multiplied by the integrand, which explains why we feel free to move it around like a multiplicative factor. – David Z Feb 9 '14 at 0:28
It's not dangerous to think of it as multiplication. It is multiplication. The dx in Leibniz notation stands for an infinitesimally small number (not equal to zero, but smaller in absolute value than any real number). For those who want some formal justification, the transfer principle from non-standard analysis says that you can manipulate infinitesimals in any way that is valid according to the elementary axioms of the real number system. ("Elementary" excludes the completeness axiom, which infinitesimals don't obey.) – Ben Crowell Feb 9 '14 at 1:33

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.

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