# How do you integrate (5x^2-9x)/((x-4)(x-1)^2) using partial fractions?

Oct 8, 2016

$\int \frac{5 {x}^{2} - 9 x}{\left(x - 4\right) {\left(x - 1\right)}^{2}} \mathrm{dx} = \frac{44}{9} \ln \left\mid x - 4 \right\mid - \frac{4}{3 \left(x - 1\right)} + \frac{1}{9} \ln \left\mid x - 1 \right\mid + C$

#### Explanation:

$\frac{5 {x}^{2} - 9 x}{\left(x - 4\right) {\left(x - 1\right)}^{2}} = \frac{A}{x - 4} + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1}$

$\textcolor{w h i t e}{\frac{5 {x}^{2} - 9 x}{\left(x - 4\right) {\left(x - 1\right)}^{2}}} = \frac{A {\left(x - 1\right)}^{2} + B \left(x - 4\right) + C \left(x - 4\right) \left(x - 1\right)}{\left(x - 4\right) {\left(x - 1\right)}^{2}}$

$\textcolor{w h i t e}{\frac{5 {x}^{2} - 9 x}{\left(x - 4\right) {\left(x - 1\right)}^{2}}} = \frac{A \left({x}^{2} - 2 x + 1\right) + B \left(x - 4\right) + C \left({x}^{2} - 5 x + 4\right)}{\left(x - 4\right) {\left(x - 1\right)}^{2}}$

$\textcolor{w h i t e}{\frac{5 {x}^{2} - 9 x}{\left(x - 4\right) {\left(x - 1\right)}^{2}}} = \frac{\left(A + C\right) {x}^{2} + \left(- 2 A + B - 5 C\right) x + \left(A - 4 B + 4 C\right)}{\left(x - 4\right) {\left(x - 1\right)}^{2}}$

Equating coefficients we get this system of linear equations:

$\left\{\begin{matrix}A + C = 5 \\ - 2 A + B - 5 C = - 9 \\ A - 4 B + 4 C = 0\end{matrix}\right.$

Adding all three equations, we find:

$- 3 B = - 4$

So $\textcolor{b l u e}{B = \frac{4}{3}}$

Adding twice the first equation to the second, we get:

$B - 3 C = 1$

Hence:

$3 C = B - 1 = \frac{4}{3} - 1 = \frac{1}{3}$

So $\textcolor{b l u e}{C = \frac{1}{9}}$

Then from the first equation:

$\textcolor{b l u e}{A =} 5 - C = 5 - \frac{1}{9} = \textcolor{b l u e}{\frac{44}{9}}$

So:

$\int \frac{5 {x}^{2} - 9 x}{\left(x - 4\right) {\left(x - 1\right)}^{2}} \mathrm{dx} = \int \left(\frac{44}{9} \cdot \frac{1}{x - 4} + \frac{4}{3} \cdot \frac{1}{x - 1} ^ 2 + \frac{1}{9} \cdot \frac{1}{x - 1}\right) \mathrm{dx}$

$\textcolor{w h i t e}{\int \frac{5 {x}^{2} - 9 x}{\left(x - 4\right) {\left(x - 1\right)}^{2}} \mathrm{dx}} = \frac{44}{9} \ln \left\mid x - 4 \right\mid - \frac{4}{3 \left(x - 1\right)} + \frac{1}{9} \ln \left\mid x - 1 \right\mid +$C