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

1 Answer
Mar 26, 2017

# 2/3x^3+7x+7ln|(x-2)/(x+2)|+1/4ln|x|-1/8ln|x-2|-1/8ln|x+2|+C,#

or,

# 2/3x^3+7x+1/4ln|x|+55/8ln|x-2|-57/8ln|x+2|+C.#

Explanation:

The Integrand, #(2x^5-x^3-1)/(x^3-4x)# is Improper Rational,

#because," 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:

#2x^5-x^3-1=ul(2x^5-8x^3)+ul(7x^3-28x)+28x-1#

#=2x^3(x^2-4)+7x(x^2-4)+28x-1.#

#:. (2x^5-x^3-1)/(x^3-4x)={2x^3(x^2-4)+7x(x^2-4)+28x-1}/{x(x^2-4)}#

#={2x^3(x^2-4)}/{x(x^2-4)}+{7x(x^2-4)}/{x(x^2-4)}+(28x)/{x(x^2-4)}-1/{x(x^2-4)}#

#=2x^2+7+28/(x^2-4)-1/{x(x^2-4)}#

#:. I=int(2x^5-x^3-1)/(x^3-4x)dx#

#=2intx^2dx+7int1dx+28int1/{x^2-(2)^2}dx-int1/{x(x-2)(x+2)}dx.#

#=2x^3/3+7x+28{1/(2*2)ln|(x-2)/(x+2)|}-J,#

#=2/3x^3+7x+7ln|(x-2)/(x+2)|-J,# where,

#J=int1/{x(x-2)(x+2)}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=[1/{x(x-2)}]_(x=-2)=(1/-2)(1/(-2-2))=1/8.#

Therefore, we have,

#J=-1/4int1/xdx+1/8int1/(x-2)dx+1/8int1/(x+2)dx,#

#=-1/4ln|x|+1/8ln|x-2|+1/8ln|x+2|.#

Altogether, we have,

#I=2/3x^3+7x+7ln|(x-2)/(x+2)|+1/4ln|x|-1/8ln|x-2|-1/8ln|x+2|+C,# or,

#=2/3x^3+7x+1/4ln|x|+55/8ln|x-2|-57/8ln|x+2|+C.#

Enjoy Maths.!