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

Aug 1, 2016

$\ln \left(\left\mid x - 3 \right\mid\right) + C$

#### 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:

$\int \frac{1}{u} \mathrm{du} = \ln \left(\left\mid u \right\mid\right) + C$

So here, we have:

$\int \frac{1}{x - 3} \mathrm{dx}$

We let: $u = x - 3$, implying that $\mathrm{du} = \mathrm{dx}$. Thus:

$\int \frac{1}{x - 3} \mathrm{dx} = \int \frac{1}{u} \mathrm{du} = \ln \left(\left\mid u \right\mid\right) + C = \ln \left(\left\mid x - 3 \right\mid\right) + C$