# How do you integrate int(x+1)/((x^2-2)(3x-5)) using partial fractions?

Mar 3, 2016

$\int \frac{x + 1}{\left({x}^{2} - 2\right) \left(3 x - 5\right)} = \int \frac{- \frac{8}{7} x - \frac{11}{7}}{{x}^{2} - 2} + \int \frac{\frac{24}{7}}{3 x - 5} \to - \frac{8}{7} \int \frac{x}{{x}^{2} - 2} - \frac{11}{7} \int \frac{1}{{x}^{2} - 2} + \frac{24}{7} \int \frac{1}{3 x - 5} \to - \frac{4}{7} \ln \left({x}^{2} - 2\right) - \frac{11 \sqrt{2}}{28} \ln \left(\frac{x - \sqrt{2}}{x + \sqrt{2}}\right) + \frac{8}{7} \ln \left(3 x - 5\right)$

#### Explanation:

To integrate we first have to resolve into partial fraction:
$\frac{x + 1}{\left({x}^{2} - 2\right) \left(3 x - 5\right)} = \frac{A x + B}{\left({x}^{2} - 2\right)} + \frac{C}{3 x - 5}$

$x + 1 = \left(A x + B\right) \left(3 x - 5\right) + C \left({x}^{2} - 2\right)$

$x + 1 = 3 A {x}^{2} + 3 B x - 5 A x - 5 B + C {x}^{2} - 2 C$

$0 = 3 A + C , 3 B - 5 A = 1 , - 5 B - 2 C = 1$

$A = \frac{8}{-} 7 , B = - \frac{11}{7} , C = \frac{24}{7}$

$\frac{x + 1}{\left({x}^{2} - 2\right) \left(3 x - 5\right)} = \frac{\frac{8}{-} 7 x + \frac{- 11}{7}}{\left({x}^{2} - 2\right)} + \frac{\frac{24}{7}}{3 x - 5}$
Then integrate each part.