# How do you use partial fractions to find the integral int 1/(4x^2-9)dx?

Dec 16, 2016

$\frac{1}{12} \ln | \frac{2 x - 3}{2 x + 3} | + C$

#### Explanation:

The denominator is the difference of the squares of $2 x$ and 3, so it easily factorizes as $\left(2 x - 3\right) \left(2 x + 3\right)$ giving the integrand as $\frac{1}{\left(2 x - 3\right) \left(2 x + 3\right)}$.
So the fraction is:
$\frac{A}{2 x - 3} + \frac{B}{2 x + 3}$ where $A$ and $B$ have to be determined by you`r method of choice (e.g. cover up rule, or equate coefficients, or susbstitute small values). Here is the cover-up rule:
$\frac{1}{\left(2 \times \frac{3}{2}\right) + 3} / \left(2 x - 3\right) + \frac{1}{2 \times \left(- \frac{3}{2}\right) - 3} / \left(2 x + 3\right)$
(To get $A$ you look at what value of x makes the expression under $A$ zero, cover up the $2 x - 3$ and replace $x$ with that value in whatever is still visible. Similarly $B$.)

This simplifies to $\left(\frac{1}{6}\right) \left(\frac{1}{2 x - 3} - \frac{1}{2 x + 3}\right)$. So the integral is now
$\left(\frac{1}{6}\right) \int \frac{1}{2 x - 3} - \frac{1}{2 x + 3} \mathrm{dx}$
$= \left(\frac{1}{6}\right) \left(\left(\frac{1}{2}\right) \ln | 2 x - 3 | - \left(\frac{1}{2}\right) \ln | 2 x + 3 |\right) + C$
$= \frac{1}{12} \ln | \frac{2 x - 3}{2 x + 3} | + C$

If you don't like the cover-up rule, then write:
$\frac{1}{4 {x}^{2} - 9} \equiv \frac{A}{2 x - 3} + \frac{B}{2 x + 3}$
which gives:
$1 \equiv \left(2 x + 3\right) A + \left(2 x - 3\right) B$
which, upon collecting similar powers of $x$, gives
$0 x + 1 \equiv \left(2 A + 2 B\right) x + \left(3 A - 2 B\right)$.
Hence, equating equal powers of $x$ you get:
$0 = 2 A + 2 B$ hence $A = - B$ and
$1 = 3 A - 3 B$ hence $1 =$6A$\mathmr{and}$A=1/6$,$B=-1/6#.

If you don't like equating similar powers, substitute any two different values for $x$ in the identity $1 \equiv \left(2 x + 3\right) A + \left(2 x - 3\right) B$ and solve the resulting simultaneous equations. If you don't like solving simultaneous equations, choose $x = \frac{3}{2}$ and then $x = - \frac{3}{2}$ for the two different values, a trick which is equivalent to the cover-up rule.