# How do you integrate f(x)=(x^2-2x)/((x^2-3)(x-3)(x-8)) using partial fractions?

Dec 5, 2017

$\int \frac{{x}^{2} - 2 x}{\left({x}^{2} - 3\right) \left(x - 3\right) \left(x - 8\right) \cdot \mathrm{dx}}$

=$\frac{5}{122} L n \left(x - 8\right)$+$\frac{21 + 5 \sqrt{3}}{732} \ln \left(x + \sqrt{3}\right)$-$\frac{1}{10} L n \left(x - 3\right)$-$\frac{21 - 5 \sqrt{3}}{6} \ln \left(x - \sqrt{3}\right) + C$

#### Explanation:

I decomposed into basic fractions,

$\frac{{x}^{2} - 2 x}{\left({x}^{2} - 3\right) \left(x - 3\right) \left(x - 8\right)}$

=$\frac{A x + B}{{x}^{2} - 3} + \frac{C}{x - 3} + \frac{D}{x - 8}$

After expanding denominator,

$\left(A x + B\right) \left(x - 3\right) \left(x - 8\right) + C \left({x}^{2} - 3\right) \left(x - 8\right) + D \left({x}^{2} - 3\right) \left(x - 3\right) = {x}^{2} - 2 x$

Set $x = 3$, $- 30 C = 3$, so $C = - \frac{1}{10}$

Set $x = 8$, $305 D = 48$, so $D = \frac{48}{305}$

Set $x = 0$, $24 B + 24 C + 9 D = 0$, so $B = - C - \frac{3}{8} \cdot D = \frac{5}{122}$

Set $x = 2$, $14 A + 14 B + 14 C + 4 D = - 1$, so $A = - \frac{7}{122}$

Hence,

$\int \frac{{x}^{2} - 2 x}{\left({x}^{2} - 3\right) \left(x - 3\right) \left(x - 8\right) \cdot \mathrm{dx}}$

=$- \frac{1}{122} \int \frac{\left(7 x - 5\right) \cdot \mathrm{dx}}{{x}^{2} - 3} - \frac{1}{10} \int \frac{\mathrm{dx}}{x - 3} + \frac{5}{122} \int \frac{\mathrm{dx}}{x - 8}$

=$\frac{5}{122} L n \left(x - 8\right) - \frac{1}{10} L n \left(x - 3\right) + C - \frac{1}{122} \cdot \int \frac{\left(7 x - 5\right) \cdot \mathrm{dx}}{{x}^{2} - 3}$

Now I solved last integral,

$\int \frac{\left(7 x - 5\right) \cdot \mathrm{dx}}{{x}^{2} - 3}$

=$\int \frac{\left(7 x - 5\right) \cdot \mathrm{dx}}{\left(x + \sqrt{3}\right) \left(x - \sqrt{3}\right)}$

=$\frac{21 - 5 \sqrt{3}}{6} \int \frac{\mathrm{dx}}{x - \sqrt{3}} - \frac{21 + 5 \sqrt{3}}{6} \int \frac{\mathrm{dx}}{x + \sqrt{3}}$

=$\frac{21 - 5 \sqrt{3}}{6} \ln \left(x - \sqrt{3}\right) - \frac{21 + 5 \sqrt{3}}{6} \ln \left(x + \sqrt{3}\right)$

Thus,

$\int \frac{{x}^{2} - 2 x}{\left({x}^{2} - 3\right) \left(x - 3\right) \left(x - 8\right) \cdot \mathrm{dx}}$

=$\frac{5}{122} L n \left(x - 8\right) + \frac{21 + 5 \sqrt{3}}{732} \ln \left(x + \sqrt{3}\right) - \frac{1}{10} L n \left(x - 3\right) - \frac{21 - 5 \sqrt{3}}{6} \ln \left(x - \sqrt{3}\right) + C$