Question

Hello! what`s the integral of ((2x+1)/x(x-2))dx?

Answer 1

Ok so i went down the wrong path first time but i have it now, have you thought about splitting this integral into partial fractions?

Answer 2

ok i`m really suck at that and what stoping me now is that in my examples we made it to look like this : \[\int\limits \frac{ 2 }{ x^2-1 } = \frac{ 2 }{ (x+1) (x-1) }\] from that i can do it myself but i`m really slow now :D and it`s been a wile since i did math last time, so i`m stuck on making bottom part of a fraction to have 2 ( ) going to google meanwile partial fractions

Answer 3

unless... because top part is higher then bottom and my formula is not applicable here...

Answer 4

ok so in the example you have given me above to split this into partial fractions this is the first step:\[\int\limits_{}^{}\frac{ 2 }{ x^{2}-1 } = \int\limits_{}^{}\frac{ 2 }{ (x+1)(x-1) } = \int\limits_{}^{}\frac{ A}{ (x+1) }\times \frac{ B }{ (x-1) }\]
So now you need to calculate A and B Do you know/remeber how do do that?
If now just say an I can explain

Answer 5

\(\huge \dfrac {2x+1}{x(x-2)}=\dfrac{A}{x}+\dfrac{B}{x-2}\)

^partial fractions

Answer 6

that would give you
\(\Large 2x+1 = A(x-2)+Bx \)

to find A, plug in x =0
to find B, plug in x=2
what do you get ?

Answer 7

yes from that i can work with.. will it be like that :
\[2=A*(x-1)+B*(x+1)\]
x=-1 2=A*(-2)+0 A=2/-2= -1
x=1 2=0+B*2 B=1
\[\int\limits (\frac{ -1 }{ x+1 }) + (\frac{ 1 }{ x-1 })dx=-\int\limits \frac{ 1 }{x+1 }+\int\limits \frac{ 1 }{ x-1 }dx=-\ln \left| x+1 \right| +\ln \left| x-1 \right| + R\]

what confusing me there is that how do i get can i make my original question to look like A/x + B/x-2 and go from there.. or do i have to have ()() two of thouse at the bottom?

Answer 8

aha.. so i do can go that way.. give me a sec

Answer 9

sure, take your time :)
when you get A,B, tell me, so that i can verify :)

Answer 10

x=0 0+1=A*(-2) A=-(1/2)
x=2 4+1=0 + B*2 B=5/2
so far..

Answer 11

correct! :)

Answer 12

\(\Huge \int \dfrac{-1}{2x}dx+\int \dfrac{5/2}{x-2}dx\)
can you integrate/proceed ?
ask if u get stuck...

Answer 13

then next it will look like this? :
\[\int\limits \frac{ 2x+1 }{ x(x-2) }=\int\limits (\frac{ -\frac{ 1 }{ 2 } }{ x }+\frac{ \frac{ 5 }{ 2 } }{ x-1 })dx=-\int\limits \frac{ \frac{ 1 }{ 2 } }{ x }dx+\int\limits \frac{ \frac{ 5 }{ 2 } }{ x-1 }dx\]

Answer 14

yes, good!

Answer 15

ow yeah.. x-2.. misread.. so far :D but this part gonna be tricky i think.. give me a moment

Answer 16

-ln|2x| + 5/2*ln|x-1| <--- is this final answer?

Answer 17

absolutely correct! :)
just add +c in the end :)

Answer 18

ow yeah.. damn thank you veeery much.. now i just need to do more exercises to fix it in my brain... thanks a lot.. do i give you some likes here or feedbacks to increase your rating or "thank you" is enough here?

Answer 19

i am just happy that you understood, and hope that you could solve similar problem on your own. That would suffice or me :)
and since you're new here, let me take this opportunity to welcome you,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Answer 20

thank you once more :D gonna try to do another one meanwhile, about surface area and volume... but i`ll probably gonna need some help on that too.. is it ok to continue this topic or better to create a new one?

Answer 21

it'd better i you 'Close' this post and ask new question in new post.
because i may not be online to help you, but there are many others here willing to help :)