# How to do integration with respect to $d(x^2)$?

I have managed to solve it using numerical techniques but I want to learn how to do it by using calculus:

$$\int_0^a \frac{1}{x^2 +x} d(x^2)$$

• If you let $y = x^{2}$, then the integral should become $\int \limits_{0}^{a} \frac{1}{y + \sqrt{y}}\,dy$. Oct 19, 2016 at 21:50
• The upper limit should be $a^2$.
– J.G.
Oct 19, 2016 at 21:55
• @J.G. Are you sure? It's $d(x^{2})$ which means $x^{2}$ ranges from $0$ to $a$. Setting $y = x^{2}$ shouldn't change the limits. Oct 19, 2016 at 22:47
• We have $d\left( x^2\right)=2xdx$. I suppose it's possible the OP could intend either meaning. However, when you see $\int_a^b u\frac{dx}{dx}dx$ written as $\int_a^b u dv\left( x\right)$ in a discussion of integration by parts, you'd probably assume $a,\,b$ are $x$ limits.
– J.G.
Oct 20, 2016 at 7:20

For any dependent variable $y$ that depends on a variable $x$, the differential of $y$ is $$dy=\frac{dy}{dx}dx,$$ where $\frac{dy}{dx}$ is, of course, the derivative of $y$ with respect to $x$. Thus, $d(x^2)=2x\,dx$.

• Many thanks! That was very helpful! Oct 19, 2016 at 21:57

In general one has $$d(f(x))=f'(x)\ dx\tag{*}$$

Note in particular that $d(x^2)=2x\ dx$ . Can you see how to go on?

A related topic to $(*)$ is called differential of a function.

• Thanks. I know how to go on from here. However, I dont know why it's equal to 2xdx Oct 19, 2016 at 21:51

This is the same as $$\int_0^a \frac{2x}{x^2 +x} dx = \int_0^a \frac{2}{x +1} dx =2\ln(a+1)$$

• Are you sure the OP didn't mean that $x^2$ should range from $0$ to $a,$ so that the integral would be $\int_0^{\sqrt{a}} \dots \,dx?$ (This assumes that $a\ge 0.)$ Oct 19, 2016 at 21:52
• I think that the bounds change to because we go from $x^2=0$ to $x^2=a$. Oct 19, 2016 at 21:52
• That's possible, I just haven't seen that before Oct 19, 2016 at 21:55