# How do you integrate (y^2 + 1) / (y^3 - 1) using partial fractions?

Oct 26, 2016

$\int \frac{\left({y}^{2} + 1\right) \mathrm{dy}}{{y}^{3} - 1} = \frac{2}{3} \ln \left(y - 1\right) + \ln \frac{{y}^{2} + y + 1}{6} - \frac{1}{\sqrt{3}} \arctan \left(\frac{2 y + 1}{\sqrt{3}}\right)$

#### Explanation:

Let's rewrite the denominator as ${y}^{3} - 1 = \left(y - 1\right) \left({y}^{2} + y + 1\right)$

And the decompsition into partial fractions
$\frac{{y}^{2} + 1}{{y}^{3} - 1} = \frac{A}{y - 1} + \frac{B x + C}{{y}^{2} + y + 1}$

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

So, ${y}^{2} + 1 = A \left({y}^{2} + y + 1\right) + \left(B y + C\right) \left(y - 1\right)$
Let $y = 0$ then, $1 = A - C$
Coefficients of ${y}^{2}$, $1 = A + B$
Coefficients of $y$, $0 = A - B + C$
Solving for A, B and C, we get
$A = \frac{2}{3}$, $B = \frac{1}{3}$ and $C = - \frac{1}{3}$
Then
int((y^2+1)dy)/(y^3-1)=int(2dy)/(3(y-1))+int((1/3y-1/3)dy)/(y^2+y+1

$= \int \frac{2 \mathrm{dy}}{3 \left(y - 1\right)} + \int \frac{\left(2 y + 1\right) \mathrm{dy}}{6 \left({y}^{2} + y + 1\right)} - \int \frac{\mathrm{dy}}{2 \left({y}^{2} + y + 1\right)}$
We have to find 3 integrals
$\int \frac{2 \mathrm{dy}}{3 \left(y - 1\right)} = \frac{2}{3} \ln \left(y - 1\right)$
$\int \frac{\left(2 y + 1\right) \mathrm{dy}}{6 \left({y}^{2} + y + 1\right)} = \ln \frac{{y}^{2} + y + 1}{6}$

And the last integral
$\int \frac{\mathrm{dy}}{2 \left({y}^{2} + y + 1\right)} = \frac{1}{2} \int \frac{\mathrm{dy}}{{y}^{2} + y + \frac{1}{4} + \frac{3}{4}}$
$= \frac{1}{2} \int \frac{\mathrm{dy}}{{\left(y + \frac{1}{2}\right)}^{2} + \frac{3}{4}}$
$= \frac{1}{2} \cdot \frac{4}{3} \int \frac{\mathrm{dy}}{{\left(\frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)}^{2} + 1}$
Let $u = \frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{2 y + 1}{\sqrt{3}}$
then $\mathrm{du} = 2 \frac{\mathrm{dy}}{\sqrt{3}}$ $\implies$$\mathrm{dy} = \frac{\sqrt{3} \mathrm{du}}{2}$
so $= \frac{2}{3} \int \frac{\frac{\sqrt{3}}{2} \mathrm{du}}{{u}^{2} + 1} = \frac{1}{\sqrt{3}} \int \frac{\mathrm{du}}{{u}^{2} + 1}$
$= \frac{1}{\sqrt{3}} \arctan u = \frac{1}{\sqrt{3}} \arctan \left(\frac{2 y + 1}{\sqrt{3}}\right)$
So putting it all together
$\int \frac{\left({y}^{2} + 1\right) \mathrm{dy}}{{y}^{3} - 1} = \frac{2}{3} \ln \left(y - 1\right) + \ln \frac{{y}^{2} + y + 1}{6} - \frac{1}{\sqrt{3}} \arctan \left(\frac{2 y + 1}{\sqrt{3}}\right)$