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

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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:

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

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