Can you do this integral?

Question

If you could help me with this integral, i would appreciate it. I know the answer is x-ln(x-1) but Step by step would be really nice. Thanks

The integral of x/(x-1) dx

Answer 1

Try partial fractions

x/(x-1) = 1+1/(x-1)

Integral of that is

=x + ln(x+1) + C

Answer 2

The partial fraction method is fine. Here's another way to do this:
Let u = x-1, x = u+1, dx = du
Then we get
∫ (u+1)/u du = ∫ du + ∫ du/u = u + ln(u) + C
Now substitute back
The answer is
x-1 + ln(x-1) + C .
Since C is arbitrary this is the same as
x + ln(x-1) + K,
where K = C-1.

Answer 3

You have to do the partila fraction decompostion to find a sum functions, where x only appears in the denominator.

For this function it is quite easy:

f(x)= x/(x-1) = (x-1+1)/(x-1) = 1 + 1/(x-1)

Integrating both terms you get:

F(x)= x + ln(1-x)

Answer 4

divide x by x-1
you will get 1+ (1/(x-1))
take integral of this
the resut is x+ln|x-1|+C

Answer 5

i think it will be
X+Ln(X-1)+CONST