Question

Integral of: (5x-1)/(2x^2-x)dx
I'm trying to use PF but something is not right. :c

Answer 1

I get the answer: lnx +3ln(2x-1)

Answer 2

Where the answer should be: lnx +3/2ln(2x-1) according to book n' wolf.

Answer 3

\[{5x-1\over2x^2-x}=\frac Ax+\frac B{2x-1}\]is this how you started?

Answer 4

Indeed!

Answer 5

first decompose it to
\[\int\limits_{}^{}1/x+3/2x-1dx\]

Answer 6

well then you probable made some small error\[{5x-1\over2x^2-x}=\frac Ax+\frac B{2x-1}\implies A(2x-1)+Bx=5x-1\]choose x=1/2 and x=0 and see what you get

Answer 7

then integrate to end up with
\[\ln \left| x \right|+(3/2)\ln \left| 2x-1 \right|+c\]

Answer 8

\[{5x-1\over2x^2-x}=\frac Ax+\frac B{2x-1}\implies A(2x-1)+Bx=5x-1\]\[x=0:-A=-1\implies A=1\]\[x=\frac12:\frac12B=\frac32\implies B=3\]so we will your answer it looks like, which means we just have some simplification issue I guess

Answer 9

Hmm, I still cant find the error. Your above answer is exactly how it went. Resulting in: 1/x + 3/(2x-1)

Answer 10

Isnt int of (3/(2x-1)) --> 3*ln(2x-1) or am I stupid? :x

Answer 11

you forgot the u-sub again, just like last time :)

Answer 12

I still dont get that part :D

Answer 13

\[\int{3\over 2x-1}dx\]\[u=2x-1\implies du=2dx\implies dx=\frac12du\]so we get\[\frac32\int{du\over u}\]

Answer 14

Unless you're supposed to u-sub everytime you got a constant - not 1 - infront of the x.

Answer 15

yes, you always must make sure you have the form du/u before integrating
without the u-sub you don't have the right form

Answer 16

Oh man!
I shall be more careful in the future, apparently missed something major like this. Super thanks for your help TuringTest!

Answer 17

Anytime :D