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

Mar 26, 2018

$\int \frac{{x}^{4} \mathrm{dx}}{x - 1} ^ 3 = 6 \ln \left\mid x - 1 \right\mid + \frac{x \left({x}^{3} + 4 {x}^{2} - 18 x + 12\right)}{2 {\left(x - 1\right)}^{2}} + C$

#### Explanation:

Using partial fractions would require lowering the degree to the numerator so that it is lower than the denominator. In this case it is easier to perform the substitution:

$u = x - 1$

$\mathrm{du} = \mathrm{dx}$

$\int \frac{{x}^{4} \mathrm{dx}}{x - 1} ^ 3 = \int \frac{{\left(u + 1\right)}^{4} \mathrm{du}}{u} ^ 3$

expanding the power of the binomial:

$\int \frac{{x}^{4} \mathrm{dx}}{x - 1} ^ 3 = \int \frac{{\left({u}^{4} + 4 {u}^{3} + 6 {u}^{2} + 4 u + 1\right)}^{4} \mathrm{du}}{u} ^ 3$

and applying linearity:

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

$\int \frac{{x}^{4} \mathrm{dx}}{x - 1} ^ 3 = {u}^{2} / 2 + 4 u + 6 \ln \left\mid u \right\mid - \frac{4}{u} - \frac{1}{2 {u}^{2}} + C$

$\int \frac{{x}^{4} \mathrm{dx}}{x - 1} ^ 3 = 6 \ln \left\mid u \right\mid + \frac{{u}^{4} + 8 {u}^{3} - 8 u - 1}{2 {u}^{2}} + C$

$\int \frac{{x}^{4} \mathrm{dx}}{x - 1} ^ 3 = 6 \ln \left\mid x - 1 \right\mid + \frac{{\left(x - 1\right)}^{4} + 8 {\left(x - 1\right)}^{3} - 8 \left(x - 1\right) - 1}{2 {\left(x - 1\right)}^{2}} + C$

$\int \frac{{x}^{4} \mathrm{dx}}{x - 1} ^ 3 = 6 \ln \left\mid x - 1 \right\mid + \frac{{x}^{4} - 4 {x}^{3} + 6 {x}^{2} - 4 x + 1 + 8 {x}^{3} - 24 {x}^{2} + 24 x - 8 - 8 x + 8 - 1}{2 {\left(x - 1\right)}^{2}} + C$

$\int \frac{{x}^{4} \mathrm{dx}}{x - 1} ^ 3 = 6 \ln \left\mid x - 1 \right\mid + \frac{{x}^{4} + 4 {x}^{3} - 18 {x}^{2} + 12 x}{2 {\left(x - 1\right)}^{2}} + C$

$\int \frac{{x}^{4} \mathrm{dx}}{x - 1} ^ 3 = 6 \ln \left\mid x - 1 \right\mid + \frac{x \left({x}^{3} + 4 {x}^{2} - 18 x + 12\right)}{2 {\left(x - 1\right)}^{2}} + C$