# How do you integrate x^2/((x-3)^2(x+4)) using partial fractions?

Jun 21, 2018

$\int {x}^{2} / \left({\left(x - 3\right)}^{2} \left(x + 4\right)\right) \mathrm{dx} = \frac{16}{49} \ln | x + 4 | + \frac{33}{49} \ln | x - 3 | - \frac{9}{7 \left(x - 3\right)} + C$

#### Explanation:

${x}^{2} / \left({\left(x - 3\right)}^{2} \left(x + 4\right)\right) = \frac{A}{x + 4} + \frac{B}{x - 3} + \frac{C}{x - 3} ^ 2$
${x}^{2} = A {\left(x - 3\right)}^{2} + B \left(x - 3\right) \left(x + 4\right) + C \left(x + 4\right)$

Substituting $x = 3$:
$9 = 7 C$ so $C = \frac{9}{7}$

Substituting $x = - 4$:
$16 = 49 A$ so $A = \frac{16}{49}$

Looking at the coefficients of ${x}^{2}$ on both sides, we see that ${x}^{2} = A {x}^{2} + B {x}^{2}$ or that $A + B = 1$.
$B = 1 - \frac{16}{49} = \frac{33}{49}$

Therefore, the partial fraction decomposition of ${x}^{2} / \left({\left(x - 3\right)}^{2} \left(x + 4\right)\right)$ is $\frac{\frac{16}{49}}{x + 4} + \frac{\frac{33}{49}}{x - 3} + \frac{\frac{9}{7}}{x - 3} ^ 2$.

$\int {x}^{2} / \left({\left(x - 3\right)}^{2} \left(x + 4\right)\right) \mathrm{dx} = \int \left(\frac{\frac{16}{49}}{x + 4} + \frac{\frac{33}{49}}{x - 3} + \frac{\frac{9}{7}}{x - 3} ^ 2\right) \mathrm{dx}$

$\int \left(\frac{\frac{16}{49}}{x + 4} + \frac{\frac{33}{49}}{x - 3} + \frac{\frac{9}{7}}{x - 3} ^ 2\right) \mathrm{dx} = \frac{16}{49} \ln | x + 4 | + \frac{33}{49} \ln | x - 3 | - \frac{9}{7 \left(x - 3\right)} + C$

Jun 21, 2018

$\int \frac{{x}^{2} \mathrm{dx}}{{\left(x - 3\right)}^{2} \left(x + 4\right)} = \frac{33}{49} \ln \left\mid x - 3 \right\mid - \frac{9}{7 \left(x - 3\right)} + \frac{16}{49} \ln \left\mid x + 4 \right\mid + C$

#### Explanation:

Apply partial fraction decomposition:

${x}^{2} / \left({\left(x - 3\right)}^{2} \left(x + 4\right)\right) = \frac{A}{x - 3} + \frac{B}{x - 3} ^ 2 + \frac{C}{x + 4}$

${x}^{2} / \left({\left(x - 3\right)}^{2} \left(x + 4\right)\right) = \frac{A \left(x - 3\right) \left(x + 4\right) + B \left(x + 4\right) + C {\left(x - 3\right)}^{2}}{{\left(x - 3\right)}^{2} \left(x + 4\right)}$

x^2 = A(x^2+x-12)+Bx+4B+C(x^2-6x+9))

${x}^{2} = A {x}^{2} + A x - 12 A + B x + 4 B + C {x}^{2} - 6 C x + 9 C$

${x}^{2} = \left(A + C\right) {x}^{2} + \left(A + B - 6 C\right) x - \left(12 A - 4 B - 9 C\right)$

$\left\{\begin{matrix}A + C = 1 \\ A + B - 6 C = 0 \\ 12 A - 4 B - 9 C = 0\end{matrix}\right.$

$\left\{\begin{matrix}A = 1 - C \\ B - 7 C = - 1 \\ 4 B + 21 C = 12\end{matrix}\right.$

$\left\{\begin{matrix}A = 1 - C \\ - 4 B + 28 C = 4 \\ 4 B + 21 C = 12\end{matrix}\right.$

$\left\{\begin{matrix}A = \frac{33}{49} \\ B = \frac{9}{7} \\ C = \frac{16}{49}\end{matrix}\right.$

Then:

$\int \frac{{x}^{2} \mathrm{dx}}{{\left(x - 3\right)}^{2} \left(x + 4\right)} = \frac{33}{49} \int \frac{\mathrm{dx}}{x - 3} + \frac{9}{7} \int \frac{\mathrm{dx}}{x - 3} ^ 2 + \frac{16}{49} \int \frac{\mathrm{dx}}{x + 4}$

$\int \frac{{x}^{2} \mathrm{dx}}{{\left(x - 3\right)}^{2} \left(x + 4\right)} = \frac{33}{49} \ln \left\mid x - 3 \right\mid - \frac{9}{7 \left(x - 3\right)} + \frac{16}{49} \ln \left\mid x + 4 \right\mid + C$