partial fractions?

Question

solve intergral of ((3x)/(x2-6x+9))dx using partial fractions to determine the antiderivitive that passes through the point (4,0)

i'm stuck

Answer 1

First break the expression apart into partial fractions.

3x / (x² - 6x + 9) = 3x / (x - 3)² = a/(x - 3) + b/(x - 3)²

Multiply thru by (x - 3)².

3x = a(x - 3) + b
3x + 0 = ax - 3a + b

Now we have two equations.

3x = ax
0 = -3a + b

3 = a
3a = b

b = 3*3 = 9

So we have:

3x / (x² - 6x + 9) = a/(x - 3) + b/(x - 3)²
3x / (x² - 6x + 9) = 3/(x - 3) + 9/(x - 3)²

Now we can integrate.

∫[3x / (x² - 6x + 9)] dx

= ∫[3/(x - 3) + 9/(x - 3)²]dx

= 3ln(x - 3) - 9/(x - 3) + C

Now solve for C.

0 = 3ln(4 - 3) - 9/(4 - 3) + C
0 = 3ln(1) - 9/1 + C
0 = 3*0 - 9 + C
C = 9

The answer then is:

3ln(x - 3) - 9/(x - 3) + 9