# How do you integrate x^2/((x-3)(x+2)^2)?

Feb 13, 2017

The answer is $= \frac{9}{25} \ln \left(| x - 3 |\right) + \frac{16}{25} \ln \left(| x + 2 |\right) + \frac{4}{5 \left(x + 2\right)} + C$

#### Explanation:

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

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

Let $x = 3$, $\implies$, $9 = 25 A$, $\implies$, $A = \frac{9}{25}$

Let $x = - 2$, $\implies$, $4 = - 5 B$, $\implies$, B=-4/5

Coefficients of ${x}^{2}$,

$1 = A + C$

$C = 1 - A = 1 - \frac{9}{25} = \frac{16}{25}$

Therefore,

${x}^{2} / \left(\left(x - 3\right) {\left(x + 2\right)}^{2}\right) = \frac{\frac{9}{25}}{x - 3} + \frac{- \frac{4}{5}}{x + 2} ^ 2 + \frac{\frac{16}{25}}{x + 2}$

So,

$\int \frac{{x}^{2} \mathrm{dx}}{\left(x - 3\right) {\left(x + 2\right)}^{2}} = \frac{9}{25} \int \frac{\mathrm{dx}}{x - 3} - \frac{4}{5} \int \frac{\mathrm{dx}}{x + 2} ^ 2 + \frac{16}{25} \int \frac{\mathrm{dx}}{x + 2}$

$= \frac{9}{25} \ln \left(| x - 3 |\right) + \frac{16}{25} \ln \left(| x + 2 |\right) + \frac{4}{5 \left(x + 2\right)} + C$