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

Dec 30, 2017

The answer is $= - \frac{1}{2 {\left(x - 1\right)}^{2}} - \frac{4}{x - 1} + 6 \ln \left(| x - 1 |\right) + {x}^{2} / 2 + 3 x + C$

#### Explanation:

The degree of the numerator is greater than the degree of the denominator

Perform a polynomial long division

${x}^{4} / {\left(x - 1\right)}^{3} = \frac{6 {x}^{2} - 8 x + 3}{x - 1} ^ 3 + x + 3$

Now, we can perform the decomposition into partial fractions

$\frac{6 {x}^{2} - 8 x + 3}{x - 1} ^ 3 = \frac{A}{x - 1} ^ 3 + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1}$

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

The denominators are the same, compare the numerators

$\left(6 {x}^{2} - 8 x + 3\right) = A + B \left(x - 1\right) + C {\left(x - 1\right)}^{2}$

Let $x = 1$, $\implies$, $A = 1$

Let $x = 0$, $\implies$, $3 = A - B + C$

$\implies$, $C - B = 2$

Coefficients of ${x}^{2}$

$6 = C$

$B = 6 - 2 = 4$

Therefore,

$\frac{6 {x}^{2} - 8 x + 3}{x - 1} ^ 3 = \frac{1}{x - 1} ^ 3 + \frac{4}{x - 1} ^ 2 + \frac{6}{x - 1}$

So,

${x}^{4} / {\left(x - 1\right)}^{3} = \frac{1}{x - 1} ^ 3 + \frac{4}{x - 1} ^ 2 + \frac{6}{x - 1} + x + 3$

Finally,

$\int \frac{{x}^{4} \mathrm{dx}}{x - 1} ^ 3 = \int \frac{\mathrm{dx}}{x - 1} ^ 3 + 4 \int \frac{\mathrm{dx}}{x - 1} ^ 2 + 6 \int \frac{\mathrm{dx}}{x - 1} + \int x \mathrm{dx} + \int 3 \mathrm{dx}$

$= - \frac{1}{2 {\left(x - 1\right)}^{2}} - \frac{4}{x - 1} + 6 \ln \left(| x - 1 |\right) + {x}^{2} / 2 + 3 x + C$