Question

What is the integral of `int (x-2)/(x-1)` from 0 to 2?

Answer 1

`2`

Explanation:

We can rewrite `x-2` as `x-1-1`. We do this so that it mimics the base.

`int_0^2(x-2)/(x-1)dx=int_0^2(x-1-1)/(x-1)dx`

`=int_0^2((x-1)/(x-1)-1/(x-1))dx=int_0^2(1-1/(x-1))dx`

All I've shown here is that `(x-2)/(x-1)=1-1/(x-1)`. You could also show this through polynomial long division or synthetic division.

To integrate this, I will split it into two indefinite integrals (for now). Later, we will come back and evaluate the combined integral from `0` to `2`.

We have:

`int1dx=x+C`

And, slightly trickier:

`-int1/(x-1)=-lnabs(x-1)+C`

For the previous integral, notice that the derivative of the denominator is present in the numerator. This means that we have a natural logarithm integral present.

Combining these, we want to evaluate

`=[x-lnabs(x-1)color(white)(""/"")]_0^2`

`=(2-lnabs(1))-(0-lnabs(-1))`

`=(2-0)-(0-0)=2`