integrating (1/(x^2 - 4x))dx
The root of x^2 - 4x is 4, then
x^2 - 4x = (x-4)(x-4) = (x-4)^2
So
(1/(x^2 - 4x))dx = (1/((x-4)^2)dx
Making u = x-4, du/dx=1 and dx=du, then
int (1/(x^2 - 4x))dx = int(1/(u^2))du
int(1/(u^2))du = -1/u +c = 1/(x+4)+c
But this result doen't match with the right one after doing a numerical test. The correct answer is:
(1/4)*ln((x-4)/x)) +c.
What's wrong in the first way?

Question

integrating  (1/(x^2 - 4x))dx 
The root of x^2 - 4x is 4, then 
x^2 - 4x = (x-4)(x-4) = (x-4)^2
So
(1/(x^2 - 4x))dx = (1/((x-4)^2)dx 
Making u = x-4, du/dx=1 and dx=du, then
int (1/(x^2 - 4x))dx = int(1/(u^2))du
int(1/(u^2))du = -1/u +c = 1/(x+4)+c
But this result doen't match with the right one after doing a numerical test. The correct answer is:
(1/4)*ln((x-4)/x)) +c. 
What's wrong in the first way?

anonymous · Answer

this line is incorrect 
$$x^2 - 4x = (x-4)(x-4) = (x-4)^2$$

anonymous · Answer

$$(x-4)^2=x^2-8x+16$$ not $x^2-4x$

anonymous · Answer

$x^2-4x=x(x-4)$ so you can use partial fractions for this one

anonymous · Answer

Thanks satellite. I reached x2−4x=x(x−4). Could you tell me why I can't use this rule:
ax^2+bx+c =a(x-x1)(x-x2) on the first polynomial? I thought it would be right if I made:
x2−4x=1*(x-4)(x-4)=(x-4)^2. What's wrong with this? Thanks again.