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.!
