# How do you integrate f(x)=(x-2)/((x^2-3)(x-3)(x-1)) using partial fractions?

Jan 5, 2016

$\int f \left(x\right) \mathrm{dx} = \frac{1}{12} \ln \left(x + \sqrt{3}\right) + \frac{1}{12} \ln \left(x - \sqrt{3}\right) + \frac{1}{12} \ln \left(x - 3\right) - \frac{1}{4} \ln \left(x - 1\right) + C$

#### Explanation:

First we need to decompose this function to a sum of partial fractions (we assume it can be done):

$\frac{x - 2}{\left(x + \sqrt{3}\right) \left(x - \sqrt{3}\right) \left(x - 3\right) \left(x - 1\right)} = \frac{\textcolor{red}{A}}{x + \sqrt{3}} + \frac{\textcolor{g r e e n}{B}}{x - \sqrt{3}} + \frac{\textcolor{b l u e}{C}}{x - 3} + \frac{\textcolor{\mathmr{and} a n \ge}{D}}{x - 1}$

but after bringing it to common denominator, the numerator equals
$\textcolor{red}{A} \left(x - \sqrt{3}\right) \left(x - 3\right) \left(x - 1\right) + \left(x + \sqrt{3}\right) \textcolor{g r e e n}{B} \left(x - 3\right) \left(x - 1\right) + \left(x + \sqrt{3}\right) \left(x - \sqrt{3}\right) \textcolor{b l u e}{C} \left(x - 1\right) + \left(x + \sqrt{3}\right) \left(x - \sqrt{3}\right) \left(x - 3\right) \textcolor{\mathmr{and} a n \ge}{D}$

Using Vieta's formulas for 3rd degree polynomials, we get
color(red)A(x^3-(4+sqrt3)x^2+(3+4sqrt3)x-3sqrt3) +color(green)B(x^3-(4-sqrt3)x^2+(3-4sqrt3)x+3sqrt3) +color(blue)C(x^3-x^2-3x+3)+color(orange)D(x^3-3x^2-3x+9)

Combining like terms:
$\left(\textcolor{red}{A} + \textcolor{g r e e n}{B} + \textcolor{b l u e}{C} + \textcolor{\mathmr{and} a n \ge}{D}\right) {x}^{3}$
-((4+sqrt3)color(red)A+(4-sqrt3)color(green)B+color(blue)C +3color(orange)D)x^2 +((3+4sqrt3)color(red)A+(3-4sqrt3)color(green)B-3color(blue)C-3color(orange)D)x +(-3sqrt3color(red)A+3sqrt3color(green)B+3color(blue)C +9color(orange)D)

but we remember it's equal to $x - 2$, so we have
{(color(red)A+color(green)B+color(blue)C+color(orange)D=0), ((4+sqrt3)color(red)A+(4-sqrt3)color(green)B+color(blue)C +3color(orange)D=0), ((3+4sqrt3)color(red)A+(3-4sqrt3)color(green)B-3color(blue)C-3color(orange)D=1), (-3sqrt3color(red)A+3sqrt3color(green)B+3color(blue)C +9color(orange)D=-2):}

Solving the system gives:
$\textcolor{red}{A} = \frac{1}{12} , \textcolor{g r e e n}{B} = \frac{1}{12} , \textcolor{b l u e}{C} = \frac{1}{12} , \textcolor{\mathmr{and} a n \ge}{D} = - \frac{1}{4}$

So we have:
$f \left(x\right) = \frac{\frac{1}{12}}{x + \sqrt{3}} + \frac{\frac{1}{12}}{x - \sqrt{3}} + \frac{\frac{1}{12}}{x - 3} - \frac{\frac{1}{4}}{x - 1}$

Integral of sum is sum of integrals, so here's integral of the first term:
$\int \frac{\frac{1}{12}}{x + \sqrt{3}} \mathrm{dx} = \frac{1}{12} \int \frac{1}{x + \sqrt{3}} \mathrm{dx}$

substituting $u = x + \sqrt{3}$ and $\mathrm{du} = \mathrm{dx}$ we have
$\frac{1}{12} \int \frac{1}{u} \mathrm{du} = \frac{1}{12} \ln u + C = \frac{1}{12} \ln \left(x + \sqrt{3}\right) + C$
The other 3 terms are done similarly.