I want to "trick" some students by showing that $$\int \frac{1}{(1-x)^2} \mathrm dx = \frac1{1-x} + C$$ in one instance and $$\int \frac{1}{(1-x)^2} \mathrm dx = \frac x{1-x} + C$$ in another.

Obviously these look like two different results due to the different numerator, but as the more astute will point out, the difference between these functions is a constant, and they therefore have the same derivative.


When solving the integral, using the substitution $u = 1-x$ we naturally arrive that the first results, but is there a way of solving the integral that "naturally" yields the second result?

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    $\begingroup$ You may try showing that $$\dfrac{1}{(1-x)^2}=1+2x+3x^2+\cdots $$ and then integrate both sides, RHS will result in $\dfrac{x}{1-x}$. But of course, this method is only valid for $|x|<1$. $\endgroup$
    – V.G
    Dec 4, 2020 at 15:18
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    $\begingroup$ Maybe another point can be put into the lesson. Do the two integrals, but "forget" to write $+C$ on them. After you are done, and show the two answers differ by a constant, you can tell them to remember $+C$ in the future, to avoid this confusion. $\endgroup$
    – GEdgar
    Dec 4, 2020 at 21:21

2 Answers 2


Why not get them to differentiate $$\frac{x}{1-x}$$ using the quotient rule:

$$\frac{\mathrm d}{\mathrm dx}\left(\frac{x}{1-x}\right) = \frac{1.(1-x) - (-1).x}{(1-x)^2}$$ which simplifies very naturally to give$$\frac{1}{(1-x)^2}.$$

This would imply $$\int \frac{1}{(1-x)^2} \mathrm dx = \frac x{1-x} + C$$

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    $\begingroup$ No, I disagree. Read his question and my answer again. My suggestion is a "natural way" to get to the second result in the question. $\endgroup$ Dec 4, 2020 at 16:10
  • $\begingroup$ Yeah, I like this one. Pretty much the first way I teach integrals is as "antiderivatives", and how you can "solve" an indefinite integral by differentiating the expected result and showing that it equals the integrand. I could phrase the "trick" as having them differentiate the second form first, and solving the integral with the $u$-sub afterwards. Thanks! $\endgroup$
    – Alec
    Dec 4, 2020 at 21:24

Put $x=\sin^2(t)$; then $dx = 2\sin(t)\cos(t)\,dt$. We have $$ \int \frac{1}{(1-x)^2}\,dx \longrightarrow \int \frac{2\sin(t)\cos(t)}{(1-\sin^2(t))^2}\,dt $$ $$ =2\int \frac{\sin(t)}{\cos^3(t)}\,dt = 2\int \tan(t)\sec^2(t)\,dt $$ $$ \begin{cases}\stackrel{y=\tan(t)}{\longrightarrow}\int 2y\,dy= \tan^2(t)+C = \frac{\sin^2(t)}{1-\sin^2(t)}+C\longrightarrow \frac{x}{1-x}+C\\ \stackrel{y=\sec(t)}{\longrightarrow}\int 2y\,dy= \sec^2(t)+C = \frac{1}{1-\sin^2(t)}+C\longrightarrow \frac{1}{1-x}+C \end{cases} $$This is a general phenomenon: antiderivatives of trig functions involving tangent and secant often look dissimilar but by the FTC they must be identical up to a constant.


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