How do you integrate f(x)=x/((x-2)(x+4)(x-7)) using partial fractions?

1 Answer
Lovecraft
Jan 10, 2016

I = 1/13ln|x-2| - 18/143ln|x+4| + 7/143ln|x-7| + c

Explanation:

Let's say that there are constant a, b and d so

a/(x-2) + b/(x+4) + d/(x-7) = x/((x-2)(x+4)(x-7))

Solving it back we have

(a(x+4)(x-7) + b(x-2)(x-7) + d(x-2)(x+4))/((x-2)(x+4)(x-7)) = x/((x-2)(x+4)(x-7))

That also means,

ax^2+4ax-7ax+28a+bx^2-9bx+14b+dx^2+2dx-8d = x

Or

x^2(a+b+d) + x(4a-7a-9b+2d) + (28a+14b-8d) = x

Or, still

a + b + d = 0
-3a-9b+2d = 1
14a+7b-4d = 0

From there, it's just a question of solving the system. Using Cramer's rule for example; solve whichever you way you think is best. For brevity's sake I won't add more to it here, if you think you could benefit from an explanation on that, just contact me.

Anyhow, solving that we see that a = 1/13, b = -18/143 and d = 7/143

From there we can just say

I = 1/13intdx/(x-2) -18/143intdx/(x+4) + 7/143intdx/(x-7)

Which are easy integrals

I = 1/13ln|x-2| - 18/143ln|x+4| + 7/143ln|x-7| + c

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