# How do you integrate 1/((x^3)(x-4)) using partial fractions?

Mar 20, 2017

$\int \frac{\mathrm{dx}}{{x}^{3} \left(x - 4\right)} = \frac{x + 2}{16 {x}^{2}} + 64 \ln \left\mid \frac{x - 4}{x} \right\mid + C$

#### Explanation:

Write the function as a sum of partial fractions with parametric numerators, using as denominators all the factors of the denominator of the rational function:

$\frac{1}{{x}^{3} \left(x - 4\right)} = \frac{A}{x} ^ 3 + \frac{B}{x} ^ 2 + \frac{C}{x} + \frac{D}{x - 4}$

Execute the sum:

$\frac{1}{{x}^{3} \left(x - 4\right)} = \frac{A \left(x - 4\right) + B x \left(x - 4\right) + C {x}^{2} \left(x - 4\right) + D {x}^{3}}{{x}^{3} \left(x - 4\right)}$

Since the denominators are equal we can equate the numerators:

$A x - 4 A + B {x}^{2} - 4 B x + C {x}^{3} - 4 C {x}^{2} + D {x}^{3} = 1$

$\left(C + D\right) {x}^{3} + \left(B - 4 C\right) {x}^{2} + \left(A - 4 B\right) x - 4 A = 1$

Two polynomials are equal if the coefficients of the same degree in $x$ are equal, so:

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

$\left\{\begin{matrix}A = - \frac{1}{4} \\ B = \frac{A}{4} = - \frac{1}{16} \\ C = \frac{B}{4} = - \frac{1}{64} \\ D = - C = \frac{1}{64}\end{matrix}\right.$

Then:

$\frac{1}{{x}^{3} \left(x - 4\right)} = - \frac{1}{4 {x}^{3}} - \frac{1}{16 {x}^{2}} - \frac{1}{64 x} + \frac{1}{64 \left(x - 4\right)}$

and integrating:

$\int \frac{\mathrm{dx}}{{x}^{3} \left(x - 4\right)} = - \int \frac{\mathrm{dx}}{4 {x}^{3}} - \int \frac{\mathrm{dx}}{16 {x}^{2}} - \int \frac{\mathrm{dx}}{64 x} + \int \frac{\mathrm{dx}}{64 \left(x - 4\right)}$

$\int \frac{\mathrm{dx}}{{x}^{3} \left(x - 4\right)} = \frac{1}{8 {x}^{2}} + \frac{1}{16 x} - 64 \ln \left\mid x \right\mid + 64 \ln \left\mid x - 4 \right\mid + C$

$\int \frac{\mathrm{dx}}{{x}^{3} \left(x - 4\right)} = \frac{x + 2}{16 {x}^{2}} + 64 \ln \left\mid \frac{x - 4}{x} \right\mid + C$