(New page: * I can't seem to get the right answer for this one. Here's the problem: <math>\int_{0}^{1}\frac{x^3dx}{x^2+2x+1}</math> After dividing, I got this: <math>\int_0^1(x-2x)dx + \int_0^1\fr...) |
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Anyone have any tips? Thanks. [[User:Idryg|Idryg]] 18:58, 13 October 2008 (UTC) | Anyone have any tips? Thanks. [[User:Idryg|Idryg]] 18:58, 13 October 2008 (UTC) | ||
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| + | OHH... I just realized that integral is equal to the inverse tangent. Nevermind! | ||
Revision as of 14:08, 13 October 2008
- I can't seem to get the right answer for this one. Here's the problem:
$ \int_{0}^{1}\frac{x^3dx}{x^2+2x+1} $
After dividing, I got this:
$ \int_0^1(x-2x)dx + \int_0^1\frac{3x+2}{(x+1)^2} $
The first integral solves to $ -\frac{3}{2} $, And I use partial fractions on the second integral:
$ 3x+2 = (Ax) + (A+B) $. So, $ A = 3 $, and $ A + B = 2 $; solve for B: $ B = -1 $
Therefore, I get this:
$ -\frac{3}{2} + 3\int_0^1\frac{dx}{x+1} - \int_0^1\frac{dx}{(x+1)^2} $
I solve for the first integral, leaving:
$ -\frac{3}{2} + 3\ln2 - \int_0^1\frac{dx}{(x+1)^2} $
Now, I'm pretty sure I didn't solve the second integral correctly, because I never end up with the right answer.
This is how i integrated the second integral:
$ \int\frac{dx}{(x+1)^2} = \frac{1}{2(x+1)}\ln(x+1)^2 $
The answer in the book is
$ 3\ln(2) - 2 $
Anyone have any tips? Thanks. Idryg 18:58, 13 October 2008 (UTC)
OHH... I just realized that integral is equal to the inverse tangent. Nevermind!
