# How do you find the antiderivative of int 1/(x^2+3x) dx?

Dec 26, 2016

$\frac{1}{3} \ln | x | - \frac{1}{3} \ln | x + 3 | + C$.

#### Explanation:

Write down:
$= \int \frac{1}{x \left(x + 3\right)} \mathrm{dx}$
$= \int \left(\frac{}{x} + \frac{}{x + 3}\right) \mathrm{dx}$.

Then apply the cover-up rule for partial fractions. To find out what goes over the $x$, use your finger to cover up the factor $x$ in the denominator of the fraction on the first line, and replace all other $x$'s with zero. Similarly, to find out what goes over the $x + 3$, cover up the $x + 3$ and replace the other $x$ with $- 3$. In each case, you replace $x$ with whatever value of $x$ makes the expression under your finger zero.

$= \int \frac{\frac{1}{0 + 3}}{x} + \frac{\frac{1}{- 3}}{x + 3} \mathrm{dx}$
$= \frac{1}{3} \int \frac{1}{x} - \frac{1}{x + 3} \mathrm{dx}$.