Question

Question #913e7

Answer 1

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

Explanation:

We can use a simple `u` - substitution...

Let `u = x-1`

`=> u +1 = x`

`=> du = dx`

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

`=> -= int u/u + 1/u du -= int 1 +1/u du`

`=> u + ln|u| + c`

Substituting back:

`=> x-1 + ln|x-1| +c_0` as `color(blue)(c_0 -1` is just another constant...

`=> color(red)( x + ln|x-1| + c )`

`c in RR` - Just a constant

Answer 2

`int\ x/(x-1)\ dx=x+ln|x-1|+C`

Explanation:

Alternatively, we can do a bit of a trick. If we use the fact that `x-1+1` still equals `x`, we can rewrite the integral:
`int\ x/(x-1)\ dx=int\ (x-1+1)/(x-1)\ dx=int\ (x-1)/(x-1)+1/(x-1)\ dx`

`=int\ 1+1/(x-1)\ dx=x+ln|x-1|+C`

As long as you're comfortable with doing the u-substitution of `1/(x-1)` in your head, this method will probably get you the answer quite a bit quicker.