How do you integrate int 1/(x^3-6x^2+9x) using partial fractions?

1 Answer
Roy E.
Dec 21, 2016

1/9ln|x|-1/9ln|x-3|-1/(3(x-3))+C

Explanation:

The integrand may be written 1/(x(x^2-6x+9))
-=1/(x(x-3)^2)
By using the cover-up rule for partial fractions, putting x=0, to give the 1/9:
-=(1/9)(1/x)+(Ax+B)/(x-3)^2 with A and B to be found.
Multiplying through by x(x^2-6x+9) :
1-=(1/9)(x-3)^2+Ax^2+Bx so
0x^2+0x+1-=(1/9+A)x^2+(-2/3+B)x+1
For this is identity to be true the coefficients of the powers of x must be the same on the two sides. So A=-1/9, B=2/3.
This gives the integral as
1/9int1/x+(6-x)/(x-3)^2dx

=1/9ln|x|+1/9int(6-x)/(x-3)^2dx
By substituting u=x-3, dx=du, or otherwise, the second integral becomes -3/(x-3)-ln|x-3| leading to the answer.

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