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# Integration - again!!! watch

1. Hello, tis me again - with some more stinky integration !! I'm getting a tad confused my first problemo is what method to use to work out:

1) ∫[ (x+1)²/(x²+1) ] dx
I had a go at doing it by substitution and but that really wasnt looking so pretty... so yeah any suggestions about the method I could use...

2) ∫cosxsinx
I know this is easy, I can see that it is, but brain has gone into overdrive. I dont know how to set out the answer - cos you can kinda just see that the 2 parts are related... do I have to actually write that?

2. (Original post by franks)
Hello, tis me again - with some more stinky integration !! I'm getting a tad confused my first problemo is what method to use to work out:

1) ∫[ (x+1)²/(x²+1) ] dx
I had a go at doing it by substitution and but that really wasnt looking so pretty... so yeah any suggestions about the method I could use...

2) ∫cosxsinx
I know this is easy, I can see that it is, but brain has gone into overdrive. I dont know how to set out the answer - cos you can kinda just see that the 2 parts are related... do I have to actually write that?

1) ∫cosxsinx dx
= ∫½sin2x dx
= -(1/4)cos2x + c

2) ∫(x+1)²/(x²+1) dx
= ∫(x²+1+2x)/(x²+1) dx
= ∫(x²+1)/(x²+1) + 2x/(x²+1) dx
= ∫1 + 2x/(x²+1) dx
= x + ln(x²+1) + c
3. 1) ∫[ (x+1)²/(x²+1) ] dx
∫[(x²+2x+1)/(x²+1)] dx = ∫[1+2x/(x²+1)] dx NOTE d/dx(x²+1) = 2x
=x + ln(x²+1) + C

2) ∫cosxsinx
∫cosxsinx dx = ∫½sin2x dx
= -¼cos2x + C

Hope this helps
4. but how did you work it out ??? Especially ∫(x+1)²/(x²+1) dx ?! You just seem to have arrived at the answer - I have NO idea how to get there

--------------

oh dear, I'm so confused
5. (Original post by franks)
but how did you work it out ??? Especially ∫(x+1)²/(x²+1) dx ?! You just seem to have arrived at the answer - I have NO idea how to get there

--------------

oh dear, I'm so confused
see my post.
6. (Original post by franks)
but how did you work it out ??? Especially ∫(x+1)²/(x²+1) dx ?! You just seem to have arrived at the answer - I have NO idea how to get there

--------------

oh dear, I'm so confused
Expand the brackets then long division.
then manipulate the standard result of ∫1/x dx = ln(X) + C
7. I dont know how to integrate 2x/(x²+1) ... thats part of the problem - I thought you couldnt do it with ln|x| ...

--------------

oh no, i do i do !

--------------

its the one with the special rule - cos the top's the differential of the bottom whupeeee thanks for helping
8. You use the fact that ∫f'(x)/f(x)dx= ln[f(x)] + c (this comes from differentiating ln[f(x)]). In the case of 2x/(x²+1), if f(x)=(x²+1) then f'(x)=2x. Therefore ∫2x/(x²+1)dx = ln[x²+1] + c.
9. stuck again
I have absolutely no idea how to do this one - I literally have pages of working, Ive tried substitution, recognition, everything!

∫[(sec²x)/(1+tanx)³]

I also tried splitting up the bottom line and then swapping tan into sec² but that just ended up with an equation about 4 lines long
I hate this question so much!!
10. You can do it by inspection. Notice:

∫[(sec²x)/(1+tanx)³]=∫(sec²x)(1+tanx)-3
And that sec²x is the differential of (1+tanx).

Then think about what needs to be differentiated to get (sec²x)/(1+tanx)³.

Spoiler:
Show
∫[(sec²x)/(1+tanx)³]dx= -0.5(1+tanx)-2+C
11. thank youuu all sorted !

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Updated: November 23, 2005
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