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

Question

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?

Answer 1

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

Answer 2

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.