(New page: Evaluate the Integral: <math>\int \frac{dx}{2\sqrt(x)+2x}</math>. I tried setting 'u' equal to <math> 2\sqrt(x)+2x</math> and 'du' equal to <math> (\frac{1}{\sqrt(x)}+2 )dx </math>. I fa...) |
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I tried setting 'u' equal to <math> 2\sqrt(x)+2x</math> and 'du' equal to <math> (\frac{1}{\sqrt(x)}+2 )dx </math>. | I tried setting 'u' equal to <math> 2\sqrt(x)+2x</math> and 'du' equal to <math> (\frac{1}{\sqrt(x)}+2 )dx </math>. | ||
I fail to see where to go from this point. Does anyone know where to go from here? [[User:Gbrizend|Gbrizend]] | I fail to see where to go from this point. Does anyone know where to go from here? [[User:Gbrizend|Gbrizend]] | ||
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| + | I tried that too. Then I thought, hey, why not try to factor out a <math>\sqrt{x}</math> from the | ||
| + | denominator and see what happens. --[[User:Bell|Bell]] 15:20, 29 September 2008 (UTC) | ||
Revision as of 11:20, 29 September 2008
Evaluate the Integral:
$ \int \frac{dx}{2\sqrt(x)+2x} $.
I tried setting 'u' equal to $ 2\sqrt(x)+2x $ and 'du' equal to $ (\frac{1}{\sqrt(x)}+2 )dx $. I fail to see where to go from this point. Does anyone know where to go from here? Gbrizend
I tried that too. Then I thought, hey, why not try to factor out a $ \sqrt{x} $ from the denominator and see what happens. --Bell 15:20, 29 September 2008 (UTC)
