Calc Question? Integrate

I have to integrate 3 / (x-1)

I used an online tool and it said 3log(x-1). But how is this the case? Is this even right?

The actual indefinite integral is 3ln|x - 1| + C. To see why, let:
u = x - 1 ==> du = dx.

Then, we have:
∫ 3/(x - 1) dx = 3 ∫ 1/(x - 1) dx, by pulling out the 3
= 3 ∫ 1/u du, by applying substitutions
= 3ln|u| + C, by integrating (note that this is a standard integral)
= 3ln|x - 1| + C, by back-substituting u = x - 1.

By the way, for whatever reason, tools like Wolfram Alpha will not include the absolute value sign. Also, in some instances, "log" is used in place of "ln" since it is usually implied that the logarithm is to the base e, although "log" is usually used for base-10 logarithms and "ln" is usually used for base-e logarithms. Later on in Mathematics courses, you may see you use of "exp." Instead of writing:
e^(x + y),

they will write:
exp(x + y),

which has the same ambiguity with using "log" in the place of "ln;" it's clear that "exp" means "exponent," so some number is being raised to the x + y'th power, but it isn't clear what that number is. In these cases, that number is e.

I hope this helps!

The integral of 3/(x - 1) is

3 ln |x - 1|

You use Wolfram Alpha? 'Log' in the same thing as 'ln,' in that case.

∫ 3dx / (x-1)
3 ∫ dx / (x-1) constant rule
u = x - 1
du = 1dx or dx
3 ∫ du / u
3 ln| u | + c
3 ln| x - 1 | + c

3ln(x-1)