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

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

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    $\begingroup$ If you let $y = x^{2}$, then the integral should become $\int \limits_{0}^{a} \frac{1}{y + \sqrt{y}}\,dy$. $\endgroup$
    – layman
    Oct 19, 2016 at 21:50
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    $\begingroup$ The upper limit should be $a^2$. $\endgroup$
    – J.G.
    Oct 19, 2016 at 21:55
  • $\begingroup$ @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. $\endgroup$
    – layman
    Oct 19, 2016 at 22:47
  • $\begingroup$ 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. $\endgroup$
    – J.G.
    Oct 20, 2016 at 7:20

3 Answers 3

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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, $d(x^2)=2x\,dx$.

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  • $\begingroup$ Many thanks! That was very helpful! $\endgroup$
    – John_Ray24
    Oct 19, 2016 at 21:57
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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.

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  • $\begingroup$ Thanks. I know how to go on from here. However, I dont know why it's equal to 2xdx $\endgroup$
    – John_Ray24
    Oct 19, 2016 at 21:51
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This is the same as $$ \int_0^a \frac{2x}{x^2 +x} dx = \int_0^a \frac{2}{x +1} dx =2\ln(a+1) $$

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    $\begingroup$ 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 $a\ge 0.)$ $\endgroup$ Oct 19, 2016 at 21:52
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    $\begingroup$ I think that the bounds change to because we go from $x^2=0$ to $x^2=a$. $\endgroup$ Oct 19, 2016 at 21:52
  • $\begingroup$ That's possible, I just haven't seen that before $\endgroup$ Oct 19, 2016 at 21:55

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