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

Oct 10, 2016

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

#### Explanation:

$\frac{{x}^{2} + x + 1}{1 - {x}^{2}} = \frac{- {x}^{2} - x - 1}{{x}^{2} - 1}$

$\textcolor{w h i t e}{\frac{{x}^{2} + x + 1}{1 - {x}^{2}}} = \frac{- \left({x}^{2} - 1\right) - x - 2}{{x}^{2} - 1}$

$\textcolor{w h i t e}{\frac{{x}^{2} + x + 1}{1 - {x}^{2}}} = - 1 + \frac{- x - 2}{{x}^{2} - 1}$

$\textcolor{w h i t e}{\frac{{x}^{2} + x + 1}{1 - {x}^{2}}} = - 1 + \frac{- x - 2}{\left(x - 1\right) \left(x + 1\right)}$

$\textcolor{w h i t e}{\frac{{x}^{2} + x + 1}{1 - {x}^{2}}} = - 1 + \frac{A}{x - 1} + \frac{B}{x + 1}$

Use Heaviside's cover up method to find:

$A = \frac{- \left(\textcolor{b l u e}{1}\right) - 2}{\left(\textcolor{b l u e}{1}\right) + 1} = - \frac{3}{2}$

$B = \frac{- \left(\textcolor{b l u e}{- 1}\right) - 2}{\left(\textcolor{b l u e}{- 1}\right) - 1} = \frac{- 1}{- 2} = \frac{1}{2}$

So:

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

$\textcolor{w h i t e}{\int \frac{{x}^{2} + x + 1}{1 - {x}^{2}} \mathrm{dx}} = - x - \frac{3}{2} \ln \left\mid x - 1 \right\mid + \frac{1}{2} \ln \left\mid x + 1 \right\mid + C$