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Integration help

  • Thread starter Sparky_
  • Start date
  • #1
196
1
Greetings

Can you help with the following integral:

[tex] -\int \frac {2(1+x)} {1-2x-x^2} dx[/tex]

I'm reasonably sure my setup is correct up to this integral. I tried to factor and do some canceling. - no luck'

thoughts and direction

Thanks
-Sparky-
 
Last edited:

Answers and Replies

  • #2
2,063
2
Have you tried substitution?
 
  • #3
ssd
268
6
Put, x^2+2x-1=z
 
  • #4
196
1
AHHH!! thanks -

[tex] u = (1-2x-x^2) [/tex]
[tex] du = -2 - 2x dx[/tex]
[tex] dx = \frac {du} {-2(1+x)} [/tex]
[tex] -\int \frac {-du} {u} [/tex]

[tex] = ln(1-2x-x^2) [/tex]

This solution is in the exponent of "e"

and leads to the integral below.
Question: can you suggest a start for:

[tex] \int \frac {1-2x-x^2} {(x+1)^2} dx [/tex]

I've tried various substitutions again and don't see it.

I've tried [tex] u = -x^2 - 2x [/tex]
[tex] du = -2x - 2 dx [/tex]
or
[tex] (-2(x+1) )dx [/tex]
[tex] dx = \frac {du} {-2(x+1)}[/tex]

leaves me with a (x+1) term

thanks
Sparky_
 
Last edited:
  • #5
8
0
Note that [tex]1-2x-x^2 = -(x+1)^2 + 2[/tex].

Now separate, and integrate.

This is completing the square. Also you could multiply the bottom out and long divide to get a similar result.
 

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