# How do you integrate 6/((x-5)(x^2-2x+3)) using partial fractions?

May 17, 2018

$I = \frac{1}{3} \ln | x - 5 | - \frac{1}{6} \ln | {x}^{2} - 2 x + 3 | - \frac{4}{3 \sqrt{2}} {\tan}^{-} 1 \left(\frac{x - 1}{\sqrt{2}}\right)$+c

#### Explanation:

Here,

$I = \int \frac{6}{\left(x - 5\right) \left({x}^{2} - 2 x + 3\right)} \mathrm{dx}$

Partial Fractions :

color(blue)(6/((x-5)(x^2-2x+3))=A/(x-5)+(Bx+C)/(x^2-2x+3)

$\implies 6 = A \left({x}^{2} - 2 x + 3\right) + B x \left(x - 5\right) + C \left(x - 5\right)$

$\implies 6 = {x}^{2} \left(A + B\right) + x \left(- 2 A - 5 B + C\right) + 3 A - 5 C$

Comparing Coefficient of x^2, x and constant term,

$A + B = 0 \implies A = - B \to \left(1\right)$

$- 2 A - 5 B + C = 0 \to \left(2\right)$

$3 A - 5 C = 6 \to \left(3\right)$

Subst. $A = - B$ into $\left(2\right)$

$\therefore 2 B - 5 B + C = 0 \implies - 3 B + C = 0 \implies C = 3 B$

From $\left(3\right)$ we get,

$3 \left(- B\right) - 5 \left(3 B\right) = 6 \implies - 18 B = 6 \implies B = - \frac{1}{3}$

From, (1) $A = - \left(- \frac{1}{3}\right) = \frac{1}{3} \mathmr{and} C = 3 \left(- \frac{1}{3}\right) = - 1$

i.e. color(blue)( A=1/3,B=-1/3,C=-1

So

$I = \int \frac{\frac{1}{3}}{x - 5} \mathrm{dx} + \int \frac{\left(- \frac{1}{3} x\right) - 1}{{x}^{2} - 2 x + 3} \mathrm{dx}$

$= \frac{1}{3} \int \frac{1}{x - 5} \mathrm{dx} - \frac{1}{3} \int \frac{x + 3}{{x}^{2} - 2 x + 3} \mathrm{dx}$

$= \frac{1}{3} \ln | x - 5 | - \frac{1}{3} \cdot \frac{1}{2} \int \frac{2 x + 6}{{x}^{2} - 2 x + 3} \mathrm{dx}$

$= \frac{1}{3} \ln | x - 5 | - \frac{1}{6} \int \frac{2 x - 2 + 8}{{x}^{2} - 2 x + 3} \mathrm{dx}$

$= \frac{1}{3} \ln | x - 5 | - \frac{1}{6} \int \frac{2 x - 2}{{x}^{2} - 2 x + 3} \mathrm{dx} - \frac{1}{6} \int \frac{8}{{x}^{2} - 2 x + 3} \mathrm{dx}$

$= \frac{1}{3} \ln | x - 5 | - \frac{1}{6} \int \frac{\frac{d}{\mathrm{dx}} \left({x}^{2} - 2 x + 3\right)}{{x}^{2} - 2 x + 3} \mathrm{dx}$

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .}$.$- \frac{8}{6} \int \frac{1}{\left({x}^{2} - 2 x + 1 + 2\right)} \mathrm{dx}$

$= \frac{1}{3} \ln | x - 5 | - \frac{1}{6} \ln | {x}^{2} - 2 x + 3 |$
color(white)(..............................)-4/3color(red)(int1/((x-1)^2+(sqrt2)^2)dx

=1/3ln|x-5|-1/6ln|x^2-2x+3|-4/3*color(red)(1/sqrt2 tan^-1((x-1)/sqrt2)+ c

$I = \frac{1}{3} \ln | x - 5 | - \frac{1}{6} \ln | {x}^{2} - 2 x + 3 | - \frac{4}{3 \sqrt{2}} {\tan}^{-} 1 \left(\frac{x - 1}{\sqrt{2}}\right)$+c

Note :

color(red)(int1/(x^2+a^2)dx=1/atan^-1(x/a)+c