How do you use partial fraction decomposition to decompose the fraction to integrate #(16x^4)/(2x-1)^3#?

Question

1599 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-23T02:47:23

First do the division, then decompose the remainder term.

We can do a partial fraction decomposition only if the degree of the numerator is less than that of the denominator.

So for this question, we start by dividing
#(16x^4)/(2x-1)^3 = (16x^4)/(8x^3-12x^2+6x-1)#

# = 2x+3+(24x^2-16x+3)/(2x-1)^3#

We can easily integrate #2x+3#, so we now find the partial fraction decomposition of:

#(24x^2-16x+3)/(2x-1)^3# in the usual way:

#A/(2x-1) + B/(2x-1)^2 + C/(2x-1)^3 = (24x^2-16x+3)/(2x-1)^3#

Leads (after some algebra) to:

#A = 6#, #" "B=4#, and #" "C=1#

So we get:

#(16x^4)/(2x-1)^3 = 2x+3+6/(2x-1) + 4/(2x-1)^2 + 1/(2x-1)^3 #