# How do you integrate int (2x^5 -x^3 -1) / (x^3 -4x) dx using partial fractions?

Mar 26, 2017

$\frac{2}{3} {x}^{3} + 7 x + 7 \ln | \frac{x - 2}{x + 2} | + \frac{1}{4} \ln | x | - \frac{1}{8} \ln | x - 2 | - \frac{1}{8} \ln | x + 2 | + C ,$

or,

$\frac{2}{3} {x}^{3} + 7 x + \frac{1}{4} \ln | x | + \frac{55}{8} \ln | x - 2 | - \frac{57}{8} \ln | x + 2 | + C .$

#### Explanation:

The Integrand, $\frac{2 {x}^{5} - {x}^{3} - 1}{{x}^{3} - 4 x}$ is Improper Rational,

$\because , \text{ the Deg. of Nr., "5 gt 3,"the Deg. of Dr.}$ Hence, first

Long Division is required. But we can avoid it as shown below:

$2 {x}^{5} - {x}^{3} - 1 = \underline{2 {x}^{5} - 8 {x}^{3}} + \underline{7 {x}^{3} - 28 x} + 28 x - 1$

$= 2 {x}^{3} \left({x}^{2} - 4\right) + 7 x \left({x}^{2} - 4\right) + 28 x - 1.$

$\therefore \frac{2 {x}^{5} - {x}^{3} - 1}{{x}^{3} - 4 x} = \frac{2 {x}^{3} \left({x}^{2} - 4\right) + 7 x \left({x}^{2} - 4\right) + 28 x - 1}{x \left({x}^{2} - 4\right)}$

$= \frac{2 {x}^{3} \left({x}^{2} - 4\right)}{x \left({x}^{2} - 4\right)} + \frac{7 x \left({x}^{2} - 4\right)}{x \left({x}^{2} - 4\right)} + \frac{28 x}{x \left({x}^{2} - 4\right)} - \frac{1}{x \left({x}^{2} - 4\right)}$

$= 2 {x}^{2} + 7 + \frac{28}{{x}^{2} - 4} - \frac{1}{x \left({x}^{2} - 4\right)}$

$\therefore I = \int \frac{2 {x}^{5} - {x}^{3} - 1}{{x}^{3} - 4 x} \mathrm{dx}$

$= 2 \int {x}^{2} \mathrm{dx} + 7 \int 1 \mathrm{dx} + 28 \int \frac{1}{{x}^{2} - {\left(2\right)}^{2}} \mathrm{dx} - \int \frac{1}{x \left(x - 2\right) \left(x + 2\right)} \mathrm{dx} .$

$= 2 {x}^{3} / 3 + 7 x + 28 \left\{\frac{1}{2 \cdot 2} \ln | \frac{x - 2}{x + 2} |\right\} - J ,$

$= \frac{2}{3} {x}^{3} + 7 x + 7 \ln | \frac{x - 2}{x + 2} | - J ,$ where,

$J = \int \frac{1}{x \left(x - 2\right) \left(x + 2\right)} \mathrm{dx} .$

Here, we will use the Method of Partial Fraction to find $J$ as

shown below:

We let, 1/{x(x-2)(x+2)}=A/x+B/(x-2)+C/(x+2); A,B,C in RR.

$A , B , C$ can be easily determined by Heaviside's Method as,

 A=[1/{(x-2)(x+2)}]_(x=0)=-1/4;

 B=[1/{x(x+2)}]_(x=2)=(1/2)(1/(2+2))=1/8; and,

$C = {\left[\frac{1}{x \left(x - 2\right)}\right]}_{x = - 2} = \left(\frac{1}{-} 2\right) \left(\frac{1}{- 2 - 2}\right) = \frac{1}{8.}$

Therefore, we have,

$J = - \frac{1}{4} \int \frac{1}{x} \mathrm{dx} + \frac{1}{8} \int \frac{1}{x - 2} \mathrm{dx} + \frac{1}{8} \int \frac{1}{x + 2} \mathrm{dx} ,$

$= - \frac{1}{4} \ln | x | + \frac{1}{8} \ln | x - 2 | + \frac{1}{8} \ln | x + 2 | .$

Altogether, we have,

$I = \frac{2}{3} {x}^{3} + 7 x + 7 \ln | \frac{x - 2}{x + 2} | + \frac{1}{4} \ln | x | - \frac{1}{8} \ln | x - 2 | - \frac{1}{8} \ln | x + 2 | + C ,$ or,

$= \frac{2}{3} {x}^{3} + 7 x + \frac{1}{4} \ln | x | + \frac{55}{8} \ln | x - 2 | - \frac{57}{8} \ln | x + 2 | + C .$

Enjoy Maths.!