# Math Help - PARTS HELP IM NEW HERE

1. ## PARTS HELP IM NEW HERE

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

2. Originally Posted by NEWBIE
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:

$\frac 1{x^2(x - 1)} = \frac Ax + \frac B{x^2} + \frac C{x - 1}$

multiply both sides by the LCD, we get:

$1 = Ax(x - 1) + B(x - 1) + Cx^2$ ....................(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

3. $\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} = \frac{1}{x^2(x-1)}$

$Ax(x-1) + B(x-1) + Cx^2 = 1$

x = 1 ... C = 1

x = 0 ... B = -1

x = 2 ... 2A + B + 4C = 1 ... A = -1

$-\frac{1}{x} - \frac{1}{x^2} + \frac{1}{x-1} = \frac{1}{x^2(x-1)}$

now integrate

4. $

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$1 = Ax(x - 1) + B(x - 1) + Cx^2
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i got that but why did you choose 0

i did

$

$
$(a+c)x^2 + (-a+b) - b

" alt="(a+c)x^2 + (-a+b) - b

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5. 1 = Ax(x - 1) + B(x - 1) + Cx^2

i got that but why did you choose 0

i did

(a+c)x^2 + (-a+b)x - b

6. -\frac{1}{x} - \frac{1}{x^2} + \frac{1}{x-1} = \frac{1}{x^2(x-1)}

what would be the first step in integrating this

7. Originally Posted by NEWBIE

1 = Ax(x - 1) + B(x - 1) + Cx^2

i got that but why did you choose 0

i did

(a+c)x^2 + (-a+b)x - b

we chose zero to wipe out A and C and leave B. we chose 1 to wipe out A and B and leave C

8. Originally Posted by NEWBIE
-\frac{1}{x} - \frac{1}{x^2} + \frac{1}{x-1} = \frac{1}{x^2(x-1)}

what would be the first step in integrating this
do you notice that your code is not working?

please stop using color text if you don't know how. it is messing with the format

9. sorry i tried copying the code that was used

10. (a+c)x^2 + (-a+b)x - b

i set -b=1

-a+b=1

-2+c=1\

where did i go wrong

11. hello

12. Originally Posted by NEWBIE
hello
firstly, bumping is against the rules

secondly, you seem to ignore everything else we told you

13. Alternatively, you can pull out x^2 common factor and substitute $u = \frac{1}{x}$. You should get it in the form of $\int \frac{u}{1-u}~du$ which is much easier to solve.

14. lol, you're quite a character!

Here's what I would do:

$\int\frac{1}{x^3-x^2}dx$

$\int\frac{1}{x^2(x-1)}dx$

$\int\frac{1+x-x}{x^2(x-1)}dx$

$-\int\frac{x-1-x}{x^2(x-1)}dx$

$-\int\frac{x-1}{x^2(x-1)}-\frac{x}{x^2(x-1)}dx$

$-\int\frac{1}{x^2}-\frac{1}{x(x-1)}dx$

$-\int\frac{1}{x^2}+\frac{x-1-x}{x(x-1)}dx$

$-\int\frac{1}{x^2}+\frac{1}{x}-\frac{1}{x-1}dx$

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!!!