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    So I really cant get very far with this. I've been ok with substitution but with this one having a fraction in it, I'm struggling.

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    (Original post by Banny Dyrne)
    So I really cant get very far with this. I've been ok with substitution but with this one having a fraction in it, I'm struggling.

    Have you been given a substitution to use or do you need to work one out?
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    Numerator is half the differential of the denominator so the integral is 1/2log(5x^2 + 7)
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    Yeah the substitution is
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    (Original post by TommyCricket)
    Numerator is half the differential of the numerator so the integral is 1/2log(5x^2 + 7)
    Yeah I've looked at the answers and thats right, I just don't know how to get there? I've been told to use this formula

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    (Original post by Banny Dyrne)
    Yeah the substitution is
    As tommy cricket says this is not the best way to integrate this but,
    Find \frac{du}{dx} then \ 1 \div\frac{du}{dx} \ =\frac{dx}{du}
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    (Original post by Banny Dyrne)
    Yeah the substitution is
    So your integral contains a 1/u factor and you need work out du/dx which hopefully will allow you to write the 5xdx bit simply in terms of du.
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    I'm really not getting this. Here's what I've worked out so far:





    so using the formula I have...




    and thats where I get lost.
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    Now simplify..
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    ?
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    (Original post by Banny Dyrne)
    I'm really not getting this. Here's what I've worked out so far:





    so using the formula I have...




    and thats where I get lost.
    No no no no, first off you would (theoretically) be able to cancel out the x in the numerator and the denominator, but then you have a function of two dependant variables u and x, which is a pain in the arse to simplify.

    You were right in thinking:

    \frac{du}{dx}=10x

    through some nifty rearranging you have:

    {5x}\text{ dx}= \frac{1}{2}du

    Does that make things any simpler when substituting?
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    Not really! Haha. I think I'm going to go over everything from the beginning of this chapter in my textbook. I'm probably missing something so simple and I just can't get my head around it right now. Thanks for trying everyone and I'll let you know when I solve it!
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    (Original post by Banny Dyrne)
    Not really! Haha. I think I'm going to go over everything from the beginning of this chapter in my textbook. I'm probably missing something so simple and I just can't get my head around it right now. Thanks for trying everyone and I'll let you know when I solve it!
    You're substituting in u to make an integral for u.

    Would you be able to integrate

    \int{\frac{1}{2x}\text{dx}}?

    what about

    \int{\frac{1}{2u}\text{du}}?
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    Yeah,



    and

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    (Original post by Banny Dyrne)
    I'm really not getting this. Here's what I've worked out so far:





    so using the formula I have...




    and thats where I get lost.
    Imagine 5x2 + 7 and 10x are swapped. Does that make it easier to see?
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    YES! Nice one. Also thanks to everyone who's spent the last three hours encouraging me.

    Finally...
 
 
 
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