# How do you integrate (9x^2 + 1) /( x^2(x − 2)^2) using partial fractions?

Feb 28, 2017

The answer is $= - \frac{1}{4 x} - \frac{37}{4 \left(x - 2\right)} + \frac{1}{4} \ln \left(| x |\right) - \frac{1}{4} \ln \left(| x - 2 |\right) + C$

#### Explanation:

We perform the decomposition into partial fractions

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

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

As the denominators are the same, we compare the numerators

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

Let $x = 0$, $\implies$, $1 = 4 A$, $\implies$, $A = \frac{1}{4}$

Let $x = 2$, $\implies$, $37 = 4 C$, $\implies$, $C = \frac{37}{4}$

Coefficients of ${x}^{2}$

$9 = A - 4 B + C - 2 D$

Coeficients of $x$,

$0 = - 4 A + 4 B$, $\implies$, $B = A = \frac{1}{4}$

$\frac{1}{4} - 1 + \frac{37}{4} - 2 D = 9$

$2 D = \frac{37}{4} - \frac{3}{4} - 9 = \frac{34}{4} - 9 = - \frac{2}{4} = - \frac{1}{2}$

$D = - \frac{1}{4}$

So,

$\frac{9 {x}^{2} + 1}{{x}^{2} {\left(x - 2\right)}^{2}} = \frac{\frac{1}{4}}{{x}^{2}} + \frac{\frac{1}{4}}{x} + \frac{\frac{37}{4}}{x - 2} ^ 2 + \frac{- \frac{1}{4}}{x - 2}$

Therefore,

$\int \frac{\left(9 {x}^{2} + 1\right) \mathrm{dx}}{{x}^{2} {\left(x - 2\right)}^{2}} = \frac{1}{4} \int \frac{\mathrm{dx}}{{x}^{2}} + \frac{1}{4} \int \frac{\mathrm{dx}}{x} + \frac{37}{4} \int \frac{\mathrm{dx}}{x - 2} ^ 2 - \frac{1}{4} \int \frac{\mathrm{dx}}{x - 2}$

$= - \frac{1}{4 x} - \frac{37}{4 \left(x - 2\right)} + \frac{1}{4} \ln \left(| x |\right) - \frac{1}{4} \ln \left(| x - 2 |\right) + C$