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denominator and see what happens. --[[User:Bell|Bell]] 15:20, 29 September 2008 (UTC) | denominator and see what happens. --[[User:Bell|Bell]] 15:20, 29 September 2008 (UTC) | ||
| − | That did it. 'u' equals | + | That did it. 'u' equals <math> 1+ \sqrt(x)</math> and 'du' equals <math> \frac{1}{2\sqrt(x)} </math>. Both are right there in the factored equation. I tried some factoring earlier, but instead of factoring a <math> \sqrt(x) </math> from the denominator, I factored an x. That lead me nowhere. Thanks for the help Dr. Bell. [[User:Gbrizend|Gbrizend]] |
Revision as of 11:46, 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)
That did it. 'u' equals $ 1+ \sqrt(x) $ and 'du' equals $ \frac{1}{2\sqrt(x)} $. Both are right there in the factored equation. I tried some factoring earlier, but instead of factoring a $ \sqrt(x) $ from the denominator, I factored an x. That lead me nowhere. Thanks for the help Dr. Bell. Gbrizend
