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

Jun 12, 2018

$= \frac{1}{61} \left\{4 \ln | {x}^{2} - 3 | - 8 \ln | x + 8 | - \frac{\sqrt{3}}{2} \ln | \frac{x - \sqrt{3}}{x + \sqrt{3}} |\right\} + C$

#### Explanation:

Here,

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

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

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

Partial Fractions :

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

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

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

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

Comparing coefficient of ${x}^{2} , x \mathmr{and}$constant term :

$A + B = 0 \to \left(1\right) , 8 B + C = 1 \to \left(2\right) , 8 C - 3 A = 0 \to \left(3\right)$

From$\left(1\right)$ , $B = - A \to \left(4\right) \mathmr{and}$ from $\left(3\right)$ , $C = \frac{3 A}{8} \to \left(5\right)$

So, from $\left(2\right)$ , $- 8 A + \frac{3 A}{8} = 1 \implies - \left(61 A\right) = 8 \implies A = - \frac{8}{61}$

From $\left(4\right)$ , $B = \frac{8}{61}$

From $\left(5\right)$ ,$C = \frac{3}{8} \left(- \frac{8}{61}\right) \implies C = - \frac{3}{61}$

Hence,

$I = \int \left[\frac{- \frac{8}{61}}{x + 8} + \frac{\left(\frac{8}{61}\right) x - \left(\frac{3}{61}\right)}{{x}^{2} - 3}\right] \mathrm{dx}$

$= - \frac{8}{61} \int \frac{1}{x + 8} \mathrm{dx} + \frac{8}{61} \int \frac{x}{{x}^{2} - 3} \mathrm{dx} - \frac{3}{61} \int \frac{1}{{x}^{2} - 3} \mathrm{dx}$

=-8/61ln|x+8|+4/61color(red)(int(2x)/(x^2-3)dx)-3/61color(blue)(int1/(x^2- (sqrt3)^2)dx

=-8/61ln|x+8|+4/61color(red)(ln|x^2-3|)-3/61*color(blue)(1/(2sqrt3)ln|(x- sqrt3)/(x+sqrt3)|+C

$= \frac{1}{61} \left\{4 \ln | {x}^{2} - 3 | - 8 \ln | x + 8 | - \frac{\sqrt{3}}{2} \ln | \frac{x - \sqrt{3}}{x + \sqrt{3}} |\right\} + C$

Note:

color(red)((1)int(f'(x))/(f(x))dx=ln|f(x)|+c

color(blue)((2)int1/(X^2-A^2)dX=1/(2A)ln|(X-A)/(X+A)|+c

Jun 12, 2018

The answer is $= \frac{8 + \sqrt{3}}{122} \ln \left(| x + \sqrt{3} |\right) + \frac{8 - \sqrt{3}}{122} \ln \left(| x - \sqrt{3} |\right) - \frac{8}{61} \ln \left(| x + 8 |\right) + C$

#### Explanation:

The function is

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

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

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

Perform the decomposition into partial fractions

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

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

The denominators are the same, compare the numerators

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

Compare the LHS and the RHS

Coefficients of ${x}^{2}$, $\implies$, $0 = A + C$, $\implies$, $A = - C$

Let $x = - 8$, $\implies$, $- 8 = 61 C$, $\implies$, $C = - \frac{8}{61}$

$A = \frac{8}{61}$

$0 = 8 B - 3 C$, $\implies$, $B = \frac{3}{8} \cdot - \frac{8}{61} = - \frac{3}{61}$

Therefore,

$\frac{x}{\left({x}^{2} - 3\right) \left(x + 8\right)} = \frac{\frac{8}{61} x - \frac{3}{61}}{{x}^{2} - 3} + \frac{- \frac{8}{61}}{x + 8}$

The integral is

$\int \frac{x \mathrm{dx}}{\left({x}^{2} - 3\right) \left(x + 8\right)} = \frac{1}{61} \int \frac{\left(8 x - 3\right) \mathrm{dx}}{{x}^{2} - 3} - \frac{8}{61} \int \frac{\mathrm{dx}}{x + 8}$

$I = {I}_{1} - {I}_{2}$

${I}_{2} = \frac{8}{61} \int \frac{8 \mathrm{dx}}{x + 8} = \frac{8}{61} \ln \left(x + 8\right)$

For the calculation of ${I}_{2}$, perform the decomposition into partial fractions.

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

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

Comparing the numerators,

$8 x - 3 = A \left(x - \sqrt{3}\right) + B \left(x + \sqrt{3}\right)$

Let $x = - \sqrt{3}$, $\implies$, $- 8 \sqrt{3} - 3 = A \cdot - 2 \sqrt{3}$, $\implies$, $A = \frac{8 \sqrt{3} + 3}{2 \sqrt{3}} = 4 + \frac{1}{2} \sqrt{3}$

Let $x = \sqrt{3}$, $\implies$, $8 \sqrt{3} - 3 = B \cdot 2 \sqrt{3}$, $\implies$, $B = \frac{8 \sqrt{3} - 3}{2 \sqrt{3}} = 4 - \frac{1}{2} \sqrt{3}$

Therefore,

$\frac{8 x - 3}{{x}^{2} - 3} = \frac{4 + \frac{1}{2} \sqrt{3}}{x + \sqrt{3}} + \frac{4 - \frac{1}{2} \sqrt{3}}{x - \sqrt{3}}$

So,

${I}_{1} = \frac{1}{61} \int \frac{\left(8 x - 3\right) \mathrm{dx}}{{x}^{2} - 3} = \frac{1}{61} \int \frac{\left(4 + \frac{1}{2} \sqrt{3}\right) \mathrm{dx}}{x + \sqrt{3}} + \frac{1}{61} \int \frac{\left(4 - \frac{1}{2} \sqrt{3}\right) \mathrm{dx}}{x - \sqrt{3}}$

$= \frac{4 + \frac{1}{2} \sqrt{3}}{61} \ln \left(x + \sqrt{3}\right) + \frac{4 - \frac{1}{2} \sqrt{3}}{61} \ln \left(x - \sqrt{3}\right)$

$= \frac{8 + \sqrt{3}}{122} \ln \left(x + \sqrt{3}\right) + \frac{8 - \sqrt{3}}{122} \ln \left(x - \sqrt{3}\right)$

And finally,

$\int \frac{x \mathrm{dx}}{\left({x}^{2} - 3\right) \left(x + 8\right)} = \frac{8 + \sqrt{3}}{122} \ln \left(| x + \sqrt{3} |\right) + \frac{8 - \sqrt{3}}{122} \ln \left(| x - \sqrt{3} |\right) - \frac{8}{61} \ln \left(| x + 8 |\right) + C$