Integral of $(2-x)/(x-1)$

Question

So I tried doing this:

I have $$\int \frac{2-x}{x-1} \mathrm{d} x$$ I used the substitution $u = x-1$, thus $x= u+1$ and $ \mathrm du = \mathrm dx$.

So then our integral becomes $$\int \frac{2-u-1}{u} \mathrm du = \int \left(\frac{1}{u} - 1\right)\mathrm du$$ Then I integrate and get $-u + \ln u$.

But when I substitute $x-1$ for $u$ I obtain $\ln(x-1)-(x-1)$ and that isn't right.

Answer 1

$\displaystyle \frac{2-x}{x-1} = - \frac{x-1}{x-1} + \frac{1}{x-1}$

See what to do now?

Answer 2

$\frac{2-x}{x-1} = \frac{1}{x-1} -1$ so that a primitive is of the form $\ln(x-1) - x + c$, $c$ a constant

Answer 3

More explicitly, your two answers differ only by a constant. But the derivative of a constant is zero.