How to integrate -2x/ (1+x^2)

  • Thread starter sara_87
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  • #1
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Homework Statement



how do i integrate:
(-2x)/(1+x^2)^2

Homework Equations





The Attempt at a Solution


I used the substitution u=1+x^2
then i got that the integral is
1/[2(1+x^2)^2 +C

but i'm not sure if that's correct

thank you
 

Answers and Replies

  • #2
morphism
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That substitution is a good idea, but it looks like you're making a mistake afterwards. Maybe if you post your working we'll be able to show you where you went wrong.
 
  • #3
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u=1+x^2
so dx=du/2x
so =integral(1/2u^2)du
=integral{1/[2(1+x^2)^2}
and now im stuck because i realised where i went wrong
so how do i continue from here?
 
  • #4
HallsofIvy
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First the 2 in the denominator of dx= du/2x will cancel the 2 in the original integral: you should have "integral (-1/u^2)du.

(Some people find it easier not to solve for dx: du= 2x dx and you can substitute directly for the "2xdx" in the original integral. And don't forget the "-".)

More important, why in the world did you go back to the 1+ x2? Don't undo your substitution until after you have integrated!

Can you integrate [itex]\int (1/u^2) du[/itex]? (Hint: that's a power of u.)
 
  • #5
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the integral is 1/u so thats 1/(1+x^2)
is that the answer?
 
  • #6
Hootenanny
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the integral is 1/u
Not quite,

[tex]\frac{d}{du}\left(\frac{1}{u}\right) = -\frac{1}{u^2} \neq \frac{1}{u^2}[/tex]

Hint,

[tex]\int\frac{du}{u^2} = \int u^{-2} du[/tex]
 
  • #7
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[tex]\int\frac{-2x}{(1+x^2)^2}dx=\int\frac{-2x}{(1+x^2)^2}\frac{d(1+x^2)}{2x}=\int\frac{-1}{(1+x^2)^2}d(1+x^2)=-\frac{(1+x^2)^{-2+1}}{-2+1}+C=\frac{1}{1+x^2}+C[/tex]
where [tex]d(1+x^2)=2xdx[/tex]
[tex]dx=\frac{d(1+x^2)}{2x}[/tex]
 
  • #8
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no i dont have to integrate u^-2
i have to integrate -u^-2
which is 1/u ???
 
  • #9
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[tex]\int\frac{-2x}{(1+x^2)^2}dx=\int\frac{-2x}{(1+x^2)^2}\frac{d(1+x^2)}{2x}=\int\frac{-1}{(1+x^2)^2}d(1+x^2)=-\frac{(1+x^2)^{-2+1}}{-2+1}+C=\frac{1}{1+x^2}+C[/tex]
where [tex]d(1+x^2)=2xdx[/tex]
[tex]dx=\frac{d(1+x^2)}{2x}[/tex]
yes i thought that was right. thank you
 
  • #10
Hootenanny
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no i dont have to integrate u^-2
i have to integrate -u^-2
which is 1/u ???
Ahh, my mistake, you are indeed correct. Sorry :redface:
 

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