Hos do i intgrate by partial fractions?

Question

I need to find the integral of a problem like,

Integral( x/( (2x-6)(x-6)^2))

Help?

Answer 1

First set up the equation:

x/((2x-6)(x-6)²) = A/(2x-6) + B/(x-6) + C/(x-6)²

Now, multiply both sides by (2x-6)(x-6)²:

x = (x-6)² A + (2x-6)(x-6) B + (2x-6) C

Now we need to extract a system of equations we can solve for A, B, and C. There are two ways to do this -- one is to simply multiply everything out, and then equate coefficients. The other is to substitute particular values for x that reveal information about the coefficients. The second method requires some thought (as opposed to the first, which can be done algorithmically), but is usually faster. For instance, substituting x=6 reveals:

6 = 6C, so C=1

And x=3 reveals:

3 = (-3)²A, so A=1/3

And finally, using the first two and substituting x=0 reveals:

0 = (-6)² (1/3) + (-6)² B + (-6) (1)
0 = 1/3 + B - 1/6
B=-1/6

So we have that:

x/((2x-6)(x-6)²) = (1/3)/(2x-6) - (1/6)/(x-6) + 1/(x-6)²

Now we integrate this:

∫(1/3)/(2x-6) - (1/6)/(x-6) + 1/(x-6)² dx

Simplifying a bit:

∫(1/6)/(x-3) - (1/6)/(x-6) + 1/(x-6)² dx

The first part of the integral may be evaluated with the substitution u=x-3, du=dx to yield:

1/6 ln (x-3) + ∫(-1/6)/(x-6) + 1/(x-6)² dx

And the remainder may be integrated with the substitution u=x-6, du=dx to get:

1/6 ln (x-3) - 1/6 ln (x-6) - 1/(x-6) + C

And we are done.

Answer 2

When you convert that to partial fractions, youwill get something like

A / (2x - 6) + B / (x - 6) + C / (x-6)^2

The first two give logs, the second a power of (x-6)

The hard part is to find A, B, C.