i have a problem INTEGRAL [1/(X^3-X^2)] CAN SOME ONE HELP ME I GOT AS FAR AS [(A/X)+(B/X^2)+(C/X-1)] BUT I CANT SEEM TO GET THEM TO CANCEL AND GET RID OF THE DENOMINATORS LIKE I HAVE IN PREVIOUS PROBLEMS ANY HINTS OR SOLUTIONS
stop shouting.
note that you have:
$\displaystyle \frac 1{x^2(x - 1)} = \frac Ax + \frac B{x^2} + \frac C{x - 1}$
multiply both sides by the LCD, we get:
$\displaystyle 1 = Ax(x - 1) + B(x - 1) + Cx^2$ ....................(1)
let x = 0 in (1) and solve for B
then let x = 1 in (1) and solve for C
then let x = anything but 0 and 1 in (1) and solve for A (note that you will have to plug in the values you found for B and C
$\displaystyle \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} = \frac{1}{x^2(x-1)}$
$\displaystyle Ax(x-1) + B(x-1) + Cx^2 = 1$
x = 1 ... C = 1
x = 0 ... B = -1
x = 2 ... 2A + B + 4C = 1 ... A = -1
$\displaystyle -\frac{1}{x} - \frac{1}{x^2} + \frac{1}{x-1} = \frac{1}{x^2(x-1)}$
now integrate
lol, you're quite a character!
Here's what I would do:
$\displaystyle \int\frac{1}{x^3-x^2}dx$
$\displaystyle \int\frac{1}{x^2(x-1)}dx$
$\displaystyle \int\frac{1+x-x}{x^2(x-1)}dx$
$\displaystyle -\int\frac{x-1-x}{x^2(x-1)}dx$
$\displaystyle -\int\frac{x-1}{x^2(x-1)}-\frac{x}{x^2(x-1)}dx$
$\displaystyle -\int\frac{1}{x^2}-\frac{1}{x(x-1)}dx$
$\displaystyle -\int\frac{1}{x^2}+\frac{x-1-x}{x(x-1)}dx$
$\displaystyle -\int\frac{1}{x^2}+\frac{1}{x}-\frac{1}{x-1}dx$
You can choose to multiply it by -1 if you like so you don't have a minus integral.
There you go NEWBIE, now stop bumping!!!