Seemingly Simple Integration: $x/(x-1)$

Question

I am currently working on some advanced engineering math but this seemingly simple integral has me stuck. Someone please show me how to derive it. It is part of a far bigger more complex problem in differential equations I am working on. Thank you so much.

$$y = \int \dfrac{x}{x-1}\, dx$$

Answer 1

Set $u=x-1.$ Then the numerator becomes $u+1$, and $du=dx.$ You can do the rest.

Answer 2

Rewrite the integrand:

$$\int \dfrac{x}{x-1}\, dx = \int(1+ \dfrac{1}{x-1}\, )dx =x+\ln|x-1| + C$$

Answer 3

First to begin:

$$\int \frac{x}{x - 1 }dx = \int1 + \frac{1}{x-1}dx = \int 1dx + \int \frac{1}{u}dx$$

when I substitute u = x -1 so x = u + 1

which gives me:

$$ x + \ln|u| + C $$

which is equivalent to

$$ x + \ln|x - 1| + C$$