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Thread: Integration question

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    Integration question

    Integrate (x+3)dx / sqrt(x^2 - 2x + 5)

    I got up to this and then my answer was:

    Sqrt(x^2 - 2x + 5) + 4ln|(sqrt(x^2-2x+5) + x - 1) / 2| + C

    But in the answer how did they randomly get rid of a half inside of the absolute value sign?

    The answer was basically

    Sqrt(x^2 - 2x + 5) + 4ln|sqrt(x^2-2x+5) + x - 1| + C

    So they just got rid of the half somehow

    Thanks for any help
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    Re: Integration question

    How did you get the 2nd part in ln? It should be in sine.

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    Re: Integration question

    Quote Originally Posted by pikachu975 View Post
    Integrate (x+3)dx / sqrt(x^2 - 2x + 5)

    I got up to this and then my answer was:

    Sqrt(x^2 - 2x + 5) + 4ln|(sqrt(x^2-2x+5) + x - 1) / 2| + C

    But in the answer how did they randomly get rid of a half inside of the absolute value sign?

    The answer was basically

    Sqrt(x^2 - 2x + 5) + 4ln|sqrt(x^2-2x+5) + x - 1| + C

    So they just got rid of the half somehow

    Thanks for any help
    Use log laws and so log|A/2| = log|A| – log(2). The – log(2) is just a constant so can be absorbed into the +C. The net effect is you can ignore the '/2' inside the log.
    Last edited by InteGrand; 23 Feb 2017 at 3:47 PM.
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    Re: Integration question

    Hope this scrawl is readable and helpful - loads of stuff from the integration trick bag ... I didn't have room to re-sub back in each variable for the longer part...

    Edit - sorry looks like you are well in control of all the method, I just read InteGrand response, I see the confusion. Oh well that was fun!
    Attached Files Attached Files
    Last edited by weaknuclearforce; 23 Feb 2017 at 4:50 PM.
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    Re: Integration question

    Quote Originally Posted by si2136 View Post
    How did you get the 2nd part in ln? It should be in sine.
    I think I thought that at first but the square root has plus not minus so I just did trig substitution

    Quote Originally Posted by InteGrand View Post
    Use log laws and so log|A/2| = log|A| – log(2). The – log(2) is just a constant so can be absorbed into the +C. The net effect is you can ignore the '/2' inside the log.
    Thanks, didn't realise this.

    Quote Originally Posted by weaknuclearforce View Post
    Hope this scrawl is readable and helpful - loads of stuff from the integration trick bag ... I didn't have room to re-sub back in each variable for the longer part...

    Edit - sorry looks like you are well in control of all the method, I just read InteGrand response, I see the confusion. Oh well that was fun!
    Thanks for the help anyway, but just a note to integrate secx just multiply top and bottom by (secx+tanx)
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    Re: Integration question

    Of course! You can tell I didn't bother following through with t-sub haha - is this a typical Ext2 Q, seems a little on the 'lets use every 1st yr Uni trick we can' - poor HSC kids, I don't envy you!

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    Re: Integration question

    Quote Originally Posted by weaknuclearforce View Post
    Of course! You can tell I didn't bother following through with t-sub haha - is this a typical Ext2 Q, seems a little on the 'lets use every 1st yr Uni trick we can' - poor HSC kids, I don't envy you!
    It's a classic trick to integrate sec(x) (the multiply top and bottom by sec + tan).

    The question in general is mainly just slightly tedious calculations I think. The method for doing it is standard and expected to be known by 4U students. (Write numerator as a*q'(x) + b for some wisely chosen constants a and b, and go from there, recalling the technique of completing the square for integrating something like 1/(sqrt(q(x)). Here, q(x) represents the quadratic in the radical and q'(x) its derivative.)
    Last edited by InteGrand; 23 Feb 2017 at 6:02 PM.
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