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Math Help - How to solve this integral?

  1. #1
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    How to solve this integral?

    Firstly thank you everyone for all your help on my other questions, I'm beginning to get how to get to the answers but still have a bunch of problems so I'll try and post them if I can't get the answer.

    For this question I'm not even sure what method to use:


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  2. #2
    Super Member
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    Jun 2008
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    Make the sub u = \frac{1}{x} which yields:
    - \int \frac{u}{\sqrt{25u^2+1}}~du

    It should be easy to solve now.
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  3. #3
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    Hello, trigoon!

    \int\frac{dx}{x^2\sqrt{x^2+25}}

    Let: x \:=\:5\tan\theta \quad\Rightarrow\quad dx \:=\:5\sec^2\!\theta\,d\theta . . . and: . \sqrt{x^2+25} \:=\:5\sec\theta


    Substitute: . \int\frac{5\sec^2\!\theta\,d\theta}{(25\tan^2\!\th  eta\cdot5\sec\theta} \;=\;\frac{1}{25}\int\frac{\sec\theta}{\tan^2\!\th  eta}\,d\theta . = \;\frac{1}{25}\int\frac{\frac{1}{\cos\theta}}{\fra  c{\sin^2\!\theta}{\cos^2\!\theta}} \;=\;\frac{1}{25}\int\frac{\cos\theta}{\sin^2\!\th  eta}\,d\theta

    . . = \;\frac{1}{25}\int\frac{1}{\sin\theta}\cdot\frac{\  cos\theta}{\sin\theta}\,d\theta \;=\;\frac{1}{25}\int\csc\theta\cot\theta\,d\theta \;=\;-\frac{1}{25}\csc\theta + C


    Back-substitute: . \tan\theta \:=\:\frac{x}{5} \quad\Rightarrow\quad \csc\theta \:=\:\frac{\sqrt{x^2+25}}{x}


    . . Answer: . -\frac{\sqrt{x^2+25}}{25x} + C

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