Argh, I can't integrate this...

[Pennycook]

Look:

u=(x+3), du=dx, dv=(1/(x+2))dx, v=ln(x+2)

1. integral ((X+3)/(x+2))=uv-integral(v*du)=(x+3)*ln(x+2)-integral(ln(x+2)dx)

Now, the bolded part is just simple substitution and an identity:

2. u=(x+2); integral(ln(u)du) = u*ln(u) - u = (x+2)*ln(x+2) - (x+2)

Plug 2. in for the bolded part and you get:

3. (x+3)*ln(x+2) - ((x+2)*ln(x+2) - (x+2)) = ln(x+2) + x + 2 + C

THANK YOU.

EDIT: I can follow your reasoning there, so now I'm confused about the answer given by the book... ANy idea why it only gives x + ln|x+2| and misses out the +2 on the end? Web calculator did it as well... Is it because they're taking 2 to be a part of the constant of integration?

[herulach]

you need to do the last part by change of variable., set u=x+2, du=dx, and voila

edit : thatll teach me for leaving the post box open while i go get a coffee

[krazy_olie]

Going to have a crack at this will post back in a few mins, need some practise.

[Pennycook]

Just one more question...

Can you do:

3/ (x-4)(x-1) by parts? Or am I supposed to be doing partial fractions?

EDIT: Can I even DO partial fractions on that? I'm thinking no... Lol.

[krazy_olie]

That's definately a partial fractions question

[TheAnimus]

Yes its partial fractions, and now herulach has fallen for the same trap as I, i don't feal so stupid. Check that mathsnet link i first posted incorrectly, they should have some examples of partial fractoins, this will get you used to seeing the "form" that you use, there are only 3 you need worry about iirc.

[Pennycook]

Yes its partial fractions, and now herulach has fallen for the same trap as I, i don't feal so stupid. Check that mathsnet link i first posted incorrectly, they should have some examples of partial fractoins, this will get you used to seeing the "form" that you use, there are only 3 you need worry about iirc.

Yeah, I did it no problems, but thanks for the link. ^_^

I was just unsure, 'caus I've never done one with just a constant on the top before. XD

[Twigman]

THANK YOU.

EDIT: I can follow your reasoning there, so now I'm confused about the answer given by the book... ANy idea why it only gives x + ln|x+2| and misses out the +2 on the end? Web calculator did it as well... Is it because they're taking 2 to be a part of the constant of integration?

C is just a constant number...you can just incorporate the +2 in to it.

eg if it was + 2 + 3 (C=3) then you would just simplify to +5 . SO they have just taken it into account with the C.

[krazy_olie]

With regards to the first one parts is an extremely long way to go round it. Remember that integral (f(u)) dx = integral f(u) du *dx/du.

Basically set x+2 as u then fiddle around a bit, takes a minute. If you want the complete working out just pm me or something.

I'll crack on the second one.

[krazy_olie]

Should I post the answer to the partial fractions one?


Which apparently i got wrong according to some website http://www.hostsrv.com/webmab/app1/M...grate&s3=basic

Error corrected , partialised (real word?) something wrong.

[Pennycook]

Right, now this one is actually really stupid...

ye^(y^2)dy/dx = e^(2x)

I know I'm supposed to integrate the right hand side with respect to x, the left with respect to y, but it's a bitch of a term to integrate...

...Unless I take logs of both sides, or something.

[smtkr]

No, it should be easy:

integral(y*e^(y^2)dy): u=y^2; you have integral(.5*e^(u)du)=.5*e^u = .5*e^(y^2)

integral(e^(2x)dx) = .5*e^(2x)

You guys are making me feel old It's been a really really long time.

[Pennycook]

No, it should be easy:

integral(y*e^(y^2)dy): u=y^2; you have integral(.5*e^(u)du)=.5*e^u = .5*e^(y^2)

integral(e^(2x)dx) = .5*e^(2x)

You guys are making me feel old It's been a really really long time.

*Facepalm.*

I always forget to check for substitution. Thanks.

[Rave]

I'm 26. I've just learned that you can forget a lot in 8 years.

[0iD]

lol, I'm 37 & I learned you can forget prettymuch everything given another 10-odd years