# Integrating $\frac{1}{(1-x)^2}$ into two different-looking functions

Background

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.

Question

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?

• 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 $|x|<1$.
– V.G
Dec 4, 2020 at 15:18
• 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. Dec 4, 2020 at 21:21

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 $$\int \frac{1}{(1-x)^2} \mathrm dx = \frac x{1-x} + C$$
• 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!
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.