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

Nov 15, 2017

$\frac{2}{17} \cdot \ln \left({x}^{2} - 2\right) + \frac{7 \sqrt{2}}{68} \cdot L n \left(\frac{x - \sqrt{2}}{x + \sqrt{2}}\right) - \frac{4}{17} \cdot L n \left(3 x - 1\right) + C$

#### Explanation:

I decomposed integrand into basic fractions,

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

After expanding denominator,

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

Set $x = \frac{1}{3}$, $- 17 \frac{C}{9} = \frac{4}{3}$, so $C = - \frac{12}{17}$

Set $x = 0$, $- B - 2 C = 1$, so $B = - 2 C - 1 = \frac{7}{17}$

Set $x = 1$, $2 A + 2 B - C = 2$, so $A = \frac{1}{2} \cdot \left(2 + C - 2 B\right) = \frac{4}{17}$

Hence,

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

=$\frac{1}{17} \cdot \int \frac{\left(4 x + 7\right) \cdot \mathrm{dx}}{{x}^{2} - 2} - \frac{1}{17} \cdot \int \frac{12 \cdot \mathrm{dx}}{3 x - 1}$

=$\frac{2}{17} \cdot \int \frac{2 x \cdot \mathrm{dx}}{{x}^{2} - 2} + \frac{7 \sqrt{2}}{68} \cdot \int \frac{2 \sqrt{2} \cdot \mathrm{dx}}{{x}^{2} - 2} - \frac{4}{17} \cdot \int \frac{3 \cdot \mathrm{dx}}{3 x - 1}$

=$\frac{2}{17} \cdot \ln \left({x}^{2} - 2\right) + \frac{7 \sqrt{2}}{68} \cdot L n \left(\frac{x - \sqrt{2}}{x + \sqrt{2}}\right) - \frac{4}{17} \cdot L n \left(3 x - 1\right) + C$