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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]] | 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]] | ||
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| + | Just because I wanted practice with LaTeX... heres the problem solved | ||
| + | <math>\int \frac{dx}{2\sqrt(x)+2x}</math> | ||
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| + | Factoring out the <math>\sqrt(x)</math> gives: | ||
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| + | <math>\int \frac{dx}{2\sqrt(x)(1+\sqrt(x))}</math> | ||
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| + | Set <math>{u=1+\sqrt(x)}</math> | ||
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| + | Therefore <math> du=\frac{1dx}{2\sqrt(x)}</math> | ||
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| + | Substitute the du and u into the equation to give: | ||
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| + | <math>\int \frac{du}{u}</math> | ||
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| + | By definition, <math>\int \frac{du}{u}</math> is equal to: | ||
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| + | <math>\ln \left (|u| \right) + c </math> | ||
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| + | Substituting u back into the equation gives the final answer of: | ||
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| + | <math>\ln \left (|1+\sqrt(x)| \right) + c </math> | ||
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| + | I think this should be the answer [[User:Ctuchek|Chad Tuchek]] | ||
Revision as of 08:19, 1 October 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
Just because I wanted practice with LaTeX... heres the problem solved
$ \int \frac{dx}{2\sqrt(x)+2x} $
Factoring out the $ \sqrt(x) $ gives:
$ \int \frac{dx}{2\sqrt(x)(1+\sqrt(x))} $
Set $ {u=1+\sqrt(x)} $
Therefore $ du=\frac{1dx}{2\sqrt(x)} $
Substitute the du and u into the equation to give:
$ \int \frac{du}{u} $
By definition, $ \int \frac{du}{u} $ is equal to:
$ \ln \left (|u| \right) + c $
Substituting u back into the equation gives the final answer of:
$ \ln \left (|1+\sqrt(x)| \right) + c $
I think this should be the answer Chad Tuchek
