Question

How do you integrate `1/(x-3)` using partial fractions?

Answer 1

`ln(abs(x-3))+C`

Explanation:

This cannot be split up any further into partial fractions. In fact, this is likely one of the results of a partial fraction decomposition.

When a linear factor like `x-3` is in the denominator of a fraction like this, without any other variables present, we are almost sure to use the natural logarithm integral:

`int1/udu=ln(absu)+C`

So here, we have:

`int1/(x-3)dx`

We let: `u=x-3`, implying that `du=dx`. Thus:

`int1/(x-3)dx=int1/udu=ln(absu)+C=ln(abs(x-3))+C`