Question

Indefinite integral problem

Answer 1

1/(x^2-x) dx

Answer 2

I tried it there, using subsitution, wolfram says i am wrong, can't see where the problem is .

Answer 3

i think you use partial fractions...

Answer 4

Never thought of that!

Answer 5

glad to help :)

Answer 6

@lgbasallote I did partial factions there.

I got 1=A(x-1)+B(x)
Let x = 0 to cancel out B therefore A = 0

Let x = 1 to cancel out x-1 therefore B=1

Integrating 0/x + 1/(x-1) = C + ln x-1

Answer 7

Wolfram gets a completely different answer. Not sure where I went wrong.

Answer 8

A(x-1) + B(x) = 1

you used root substitution here huh...let's try Gaussian method

Ax - A + Bx = 1

constant
-A = 1
A = -1

coefficient of x
A + B = 0
-1 + B = 0
B = 1

therefore you have \[\huge \int \frac{1}{(x^2 - x)} = \int \frac{-1}{x} + \int \frac{1}{x-1}\]
\[\huge \implies -\ln x + \ln (x-1) + C\]

idk how wolfram got ln (1-x) though

Answer 9

your mistake was this

1 = A(x - 1) + B(x)

when x = 0

1 = A(0 - 1) - 0
1 = -A
A = -1

Answer 10

do you get my solution though?

Answer 11

@ironictoaster ?

Answer 12

@lgbasallote Yeah sorry I did, my internet connection went down briefly there

Answer 13

I'll try solve it now

Answer 14

good luck :D

Answer 15

ah excellent, works fine. Stupid little mistakes little this annoy me