Question

What is `f(x) = int x/(x-1) dx` if `f(2) = 0`?

Answer 1

Since `ln` can't help you, set the denominator because of its simple form as a variable. When you solve the integral, just set `x=2` to fit the `f(2)` in the equation and find the integration constant.

Answer is:

`f(x)=x+ln|x-1|-2`

Explanation:

`f(x)=intx/(x-1)dx`

The `ln` function will not help in this case. However, since the denominator is quite simple (1st grade):

Set `u=x-1=>x=u+1`

and `(du)/dx=d(x+1)/dx=(x+1)'=1=>(du)/dx=1<=>du=dx`

`intx/(x-1)dx=int(u+1)/(u)du=int(u/u+1/u)du=`

`=int(1+1/u)du=int1du+int(du)/u=u+ln|u|+c`

Substituting `x` back:

`u+ln|u|+c=x-1+ln|x-1|+c`

So:

`f(x)=intx/(x-1)dx=x-1+ln|x-1|+c`

`f(x)=x-1+ln|x-1|+c`

To find `c` we set `x=2`

`f(2)=2-1+ln|2-1|+c`

`0=1+ln1+c`

`c=-1`

Finally:

`f(x)=x-1+ln|x-1|+c=x-1+ln|x-1|-1=x+ln|x-1|-2`

`f(x)=x+ln|x-1|-2`