How do you integrate (5x-3)/((x-2)(x-3)^2) using partial fractions?

Jun 13, 2017

$\int \frac{5 x - 3}{\left(x - 2\right) {\left(x - 3\right)}^{2}} = 7 \ln | x - 2 | - 7 \ln | x - 3 | - \frac{12}{x - 3} + c$

Explanation:

Okay, this is long and I have left out lots of information. But hopefully it gives an idea of the process.
Partial fractions involves decomposing a simplified fraction back to its constituent parts. We want to split a single fraction into multiple, simpler fractions. That way we can integrate the parts separately and more easily.
We look at the terms in the denominator to see how many fractions we will end up with.
An (ax+b) term turns into an $\frac{A}{a x + b}$ term,
and an
${\left(a x + b\right)}^{2}$ term turns into two terms:
$\frac{B}{a x + b} + \frac{C}{a x + b} ^ 2$

So, our fraction will be broken up into three fractions with denominators that we know. Furthermore, we will have three unknown numerators, which we will call A, B &C. We need to solve for these.
So, following the rules just specified, we now have an equality that looks like this:

$\frac{5 x - 3}{\left(x - 2\right) {\left(x - 3\right)}^{2}} = \frac{A}{x - 2} + \frac{B}{x - 3} + \frac{C}{x - 3} ^ 2$

The next step is to get rid of the denominators by multiplying both sides by $\left(x - 2\right) {\left(x - 3\right)}^{2}$:

$\frac{\left(5 x - 3\right) \left(x - 2\right) {\left(x - 3\right)}^{2}}{\left(x - 2\right) {\left(x - 3\right)}^{2}}$

$= \frac{A \left(x - 2\right) {\left(x - 3\right)}^{2}}{x - 2} + \frac{B \left(x - 2\right) {\left(x - 3\right)}^{2}}{x - 3} + \frac{C \left(x - 2\right) {\left(x - 3\right)}^{2}}{x - 3} ^ 2$

After cancelling out, we are left with this:

$5 x - 3 = A {\left(x - 3\right)}^{2} + B \left(x - 2\right) \left(x - 3\right) + C \left(x - 2\right)$

Expand the right hand side, group all the like-terms together and factororise:

$5 x - 3 = A \left({x}^{2} - 6 x + 9\right) + B \left({x}^{2} - 5 x + 6\right) + C x - 2 C$

$0 {x}^{2} + 5 x - 3 = A {x}^{2} + B {x}^{2} - 6 A x - 5 B x + C x + 9 A + 6 B - 2 C$

$0 {x}^{2} + 5 x - 3 = {x}^{2} \left(A + B\right) + x \left(- 6 A - 5 B + C\right) + \left(9 A + 6 B - 2 C\right)$

Equate the coefficients of each term to solve. I won't show working for this part, but use substitution:

$0 = A + B$ $\Rightarrow B = - A$
$5 = - 6 A - 5 B + C$ $\Rightarrow 5 = - 6 A - 5 \left(- A\right) + C$
$- 3 = 9 A + 6 B - 2 C$

$A = 7$
$B = - 7$
$C = 12$

Finally, our decomposed expression is:

$\frac{A}{x - 2} + \frac{B}{x - 3} + \frac{C}{x - 3} ^ 2 = \frac{7}{x - 2} - \frac{7}{x - 3} + \frac{12}{{\left(x - 3\right)}^{2}}$

Now integrate each term using substitution. Treat the $\frac{\mathrm{du}}{\mathrm{dx}}$ operator just like an algebraic term and you can find an easier expression to integrate. E.g.
$\int \frac{7}{x - 2} \mathrm{dx}$
Let $u = x - 2$ $\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 1$$\Rightarrow \mathrm{du} = 1 \cdot \mathrm{dx}$

Now we can see $1 \cdot \mathrm{dx}$ in the integral so replace it with $\mathrm{du}$, then we can integrate!

$\int \frac{7}{x - 2} = 7 \int \frac{1}{u} \mathrm{dx} = 7 \int \frac{1}{u} \cdot 1 \cdot \mathrm{dx} = 7 \int \frac{1}{u} \mathrm{du}$

$7 \int \frac{1}{u} \mathrm{du} = 7 \ln | u | = 7 \ln | x - 2 |$

Same process for the second term:
$\int \frac{- 7}{x - 3} = - 7 \int \frac{1}{u} \mathrm{du} = - 7 \ln | x - 3 |$

Same method for the third term:
$\int \frac{12}{{\left(x - 3\right)}^{2}} = 12 \int \frac{1}{u} ^ 2 \mathrm{du} = 12 \int {u}^{-} 2 \mathrm{du} = - 12 {u}^{-} 1 = - \frac{12}{x - 3}$

So finaaallly:

$\int \frac{5 x - 3}{\left(x - 2\right) {\left(x - 3\right)}^{2}} = \int \frac{7}{x - 2} + \int \frac{- 7}{x - 3} + \int \frac{12}{{\left(x - 3\right)}^{2}}$

$= 7 \ln | x - 2 | - 7 \ln | x - 3 | - \frac{12}{x - 3} + c$