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<math>\int\frac{dx}{(x+1)^2} = \frac{1}{2(x+1)}\ln(x+1)^2</math> | <math>\int\frac{dx}{(x+1)^2} = \frac{1}{2(x+1)}\ln(x+1)^2</math> | ||
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The answer in the book is | The answer in the book is | ||
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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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| + | * You integrated the last integral incorrectly. To solve the last integral, substitute u for x+1 and then use simple integrating rules for powers-<math>\int u^ndu=\frac{1}{n+1}u^{n+1}</math> [[User:Jhunsber|His Awesomeness, Josh Hunsberger]] | ||
Revision as of 07:10, 14 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)
- You integrated the last integral incorrectly. To solve the last integral, substitute u for x+1 and then use simple integrating rules for powers-$ \int u^ndu=\frac{1}{n+1}u^{n+1} $ His Awesomeness, Josh Hunsberger
