# How do you integrate int x^3/((x^4-16)(x-3))dx using partial fractions?

$\int \frac{{x}^{3}}{\left({x}^{4} - 16\right) \left(x - 3\right)} = \frac{- 3}{52} \ln \left({x}^{2} + 4\right) + \frac{1}{13} {\tan}^{-} 1 \left(\frac{x}{2}\right) - \frac{1}{20} \ln \left(x + 2\right) - \frac{1}{4} \ln \left(x - 2\right) + \frac{27}{65} \ln \left(x - 3\right) + {C}_{0}$

#### Explanation:

Start from the given integral and set up the variables using A, B, C ,D, E

$\int \frac{{x}^{3}}{\left({x}^{4} - 16\right) \left(x - 3\right)} \mathrm{dx} =$

$\int \left(\frac{A x + B}{{x}^{2} + 4} + \frac{C}{x + 2} + \frac{D}{x - 2} + \frac{E}{x - 3}\right) \mathrm{dx}$

and out of the numerators from both sides we have

${x}^{3} = A {x}^{4} - 4 A {x}^{2} - 3 A {x}^{3} + 12 A x + B {x}^{3} - 4 B x - 3 B {x}^{2} + 12 B + C {x}^{4} - 5 C {x}^{3} + 10 C {x}^{2} - 20 C x + 24 C + D {x}^{4} - D {x}^{3} - 2 D {x}^{2} - 4 D x - 24 D + E {x}^{4} - 16 E$

After setting up all the equations . The following are the equations

$C + D + E = 0 \text{ }$equation 1
$B - 5 C - D = 1 \text{ }$equation 2
$- 3 B + 10 C - 2 D = 0 \text{ }$equation 3
$- 4 B - 20 C - 4 D = 0 \text{ }$equation 4
$12 B + 24 C - 24 D - 16 E = 0 \text{ }$equation 5

Solving for A, B, C, D, E by any method
$A = \frac{- 3}{26}$
$B = \frac{2}{13}$
$C = \frac{- 1}{20}$
$D = \frac{- 1}{4}$
$E = \frac{27}{65}$

We have to integrate now

$\int \frac{{x}^{3}}{\left({x}^{4} - 16\right) \left(x - 3\right)} \mathrm{dx} =$

$\int \left(\frac{\frac{- 3}{26} x + \frac{2}{13}}{{x}^{2} + 4} + \frac{- 1}{20} / \left(x + 2\right) + \frac{- 1}{4} / \left(x - 2\right) + \frac{27}{65} / \left(x - 3\right)\right) \mathrm{dx}$

$\int \frac{{x}^{3}}{\left({x}^{4} - 16\right) \left(x - 3\right)} = \frac{- 3}{52} \ln \left({x}^{2} + 4\right) + \frac{1}{13} {\tan}^{-} 1 \left(\frac{x}{2}\right) - \frac{1}{20} \ln \left(x + 2\right) - \frac{1}{4} \ln \left(x - 2\right) + \frac{27}{65} \ln \left(x - 3\right) + {C}_{0}$

where ${C}_{0}$ be the constant of integration

God bless....I hope the explanation is useful.