# How do you integrate int (x^5 + 1)/(x^6 - x^4) using partial fractions?

Oct 12, 2016

$\int \left(\frac{{x}^{5} + 1}{{x}^{6} - {x}^{4}}\right) \mathrm{dx} = {c}_{1} \log \left\mid x \right\mid + {c}_{2} / x - {c}_{3} / \left(2 {x}^{2}\right) - {c}_{4} / \left(3 {x}^{3}\right) + {c}_{4} \log \left\mid x + 1 \right\mid + {c}_{5} \log \left\mid x - 1 \right\mid + {c}_{0}$

#### Explanation:

$\frac{{x}^{5} + 1}{{x}^{6} - {x}^{4}} = \frac{{x}^{5} + 1}{{x}^{4} \left({x}^{2} - 1\right)} =$
$= {c}_{1} / x + {c}_{2} / {x}^{2} + {c}_{3} / {x}^{3} + {c}_{4} / {x}^{4} + {c}_{5} / \left(x + 1\right) + {c}_{6} / \left(x - 1\right)$

Then

$\int \left(\frac{{x}^{5} + 1}{{x}^{6} - {x}^{4}}\right) \mathrm{dx} = \int \left({c}_{1} / x + {c}_{2} / {x}^{2} + {c}_{3} / {x}^{3} + {c}_{4} / {x}^{4} + {c}_{5} / \left(x + 1\right) + {c}_{6} / \left(x - 1\right)\right) \mathrm{dx} =$

$= {c}_{1} \log \left\mid x \right\mid + {c}_{2} / x - {c}_{3} / \left(2 {x}^{2}\right) - {c}_{4} / \left(3 {x}^{3}\right) + {c}_{4} \log \left\mid x + 1 \right\mid + {c}_{5} \log \left\mid x - 1 \right\mid + C$

The ${c}_{k}$ determination follows.

Equating

$\frac{{x}^{5} + 1}{{x}^{6} - {x}^{4}} = {c}_{1} / x + {c}_{2} / {x}^{2} + {c}_{3} / {x}^{3} + {c}_{4} / {x}^{4} + {c}_{5} / \left(x + 1\right) + {c}_{6} / \left(x - 1\right)$

The condition to mantain the equality $\forall x \in \mathbb{R}$ is the observation of

$\left\{\begin{matrix}1 + {c}_{4} = 0 \\ {c}_{3} = 0 \\ {c}_{2} - {c}_{4} = 0 \\ {c}_{1} - {c}_{3} = 0 \\ {c}_{2} - {c}_{5} + {c}_{6} = 0 \\ {c}_{1} + {c}_{5} + {c}_{6} - 1 = 0\end{matrix}\right.$

Solving we obtain

$\left({c}_{1} = 0 , {c}_{2} = - 1 , {c}_{3} = 0 , {c}_{4} = - 1 , {c}_{5} = 0 , {c}_{6} = 1\right)$