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

Dec 22, 2016

The answer is =1/12ln(∣x-1∣)+1/28ln(∣x-3∣)-5/84ln(x^2+5)- 4/(21sqrt5)arctan(x/sqrt5)+C

#### Explanation:

Let's perform the decomposition into partial fractions

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

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

Therefore,

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

Let $x = 1$, $\implies$, $- 1 = - 12 A$, $\implies$, $A = \frac{1}{12}$

Let $x = 3$, $\implies$, $1 = 28 B$, $\implies$, $B = \frac{1}{28}$

Coefficients of ${x}^{3}$, $\implies$, $0 = A + B + C$

$C = - \frac{5}{42}$

Let $x = 0$, $\implies$, $- 2 = - 15 A - 5 B + 3 D$

$3 D = 15 A + 5 B - 2 = \frac{15}{12} + \frac{5}{28} - 2 = - \frac{4}{7}$

$D = - \frac{4}{21}$

Therefore,

$\frac{x - 2}{\left(x - 1\right) \left(x - 3\right) \left({x}^{2} + 5\right)} = \frac{\frac{1}{12}}{x - 1} + \frac{\frac{1}{28}}{x - 3} + \frac{- \frac{5}{42} x - \frac{4}{21}}{{x}^{2} + 5}$

$= \frac{\frac{1}{12}}{x - 1} + \frac{\frac{1}{28}}{x - 3} - \frac{1}{42} \frac{5 x + 8}{{x}^{2} + 5}$

So,

$I = \int \frac{\left(x - 2\right) \mathrm{dx}}{\left(x - 1\right) \left(x - 3\right) \left({x}^{2} + 5\right)} = \frac{1}{12} \int \frac{\mathrm{dx}}{x - 1} + \frac{1}{28} \int \frac{\mathrm{dx}}{x - 3} - \frac{1}{42} \int \frac{\left(5 x + 8\right) \mathrm{dx}}{{x}^{2} + 5}$

=1/12ln(∣x-1∣)+1/28ln(∣x-3∣)-1/42(5int(xdx)/(x^2+5)+8int(dx)/(x^2+5))

Let $u = {x}^{2} + 5$, $\implies$, $\mathrm{du} = 2 x \mathrm{dx}$

int(xdx)/(x^2+5)=1/2 int (du)/u=1/2lnu=1/2ln(∣x^2+5∣)

Let $v = \frac{x}{\sqrt{5}}$, $\implies$, $\mathrm{dv} = \frac{\mathrm{dx}}{\sqrt{5}}$

$8 \int \frac{\mathrm{dx}}{{x}^{2} + 5} = \frac{8}{\sqrt{5}} \int \frac{\mathrm{dv}}{{v}^{2} + 1}$

Let $v = \tan \theta$, $\mathrm{dv} = {\sec}^{2} \theta d \theta$

and $1 + {\tan}^{2} \theta = {\sec}^{2} \theta$

So,

$\frac{8}{\sqrt{5}} \int \frac{\mathrm{dv}}{{v}^{2} + 1} = \frac{8}{\sqrt{5}} \int \left({\sec}^{2} \frac{\theta}{\sec} ^ 2 \theta d \theta\right)$

$= \frac{8}{\sqrt{5}} \theta = \frac{8}{\sqrt{5}} \arctan v = \frac{8}{\sqrt{5}} \arctan \left(\frac{x}{\sqrt{5}}\right)$

Putting it all together

I=1/12ln(∣x-1∣)+1/28ln(∣x-3∣)-5/84ln(x^2+5)- 4/(21sqrt5)arctan(x/sqrt5)+C