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Math Help - Simple integration question

  1. #1
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    Simple integration question

    I'm looking to integrate:

    1/(2x+5)



    Wolframalpha.com displays the answer as:

    1/2 log(2x+5)



    My problem is with the (1/2) in the answer. Wolfram claims that you substitute u=2x+5 and du=2dx, but then goes on to take the 1/2 out before substituting then integrating...


    Could someone take me step by step through this?
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  2. #2
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    Re: Simple integration question

    Quote Originally Posted by remember View Post
    I'm looking to integrate:

    1/(2x+5)



    Wolframalpha.com displays the answer as:

    1/2 log(2x+5)



    My problem is with the (1/2) in the answer. Wolfram claims that you substitute u=2x+5 and du=2dx, but then goes on to take the 1/2 out before substituting then integrating...


    Could someone take me step by step through this?
    Hi remember,

    \int\frac{1}{2x+5}dx

    \mbox{Substitute, }u=2x+5\Rightarrow dx=\frac{du}{2}

    \int\frac{1}{2u}du=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}\ln|u|+C=\frac{1}{2}\ln|2  x+5|+C

    Therefore, \int\frac{1}{2x+5}dx=\frac{1}{2}\ln|2x+5|+C
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  3. #3
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    Re: Simple integration question

    Quote Originally Posted by remember View Post
    I'm looking to integrate:

    1/(2x+5)



    Wolframalpha.com displays the answer as:

    1/2 log(2x+5)



    My problem is with the (1/2) in the answer. Wolfram claims that you substitute u=2x+5 and du=2dx, but then goes on to take the 1/2 out before substituting then integrating...
    Yes, of course. If c is any constant then \int cf(x)dx= c\int f(x)dx

    Could someone take me step by step through this?
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  4. #4
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    Re: Simple integration question

    Thanks a lot! I really need to brush up on the basics!
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