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    Hi guys, I'm doing my C3 calculus work and I'm stuck on a question, it goes:

    Integrate: x(4-3x^2)^1/2 and give your answer as a fraction.

    I'm struggling to decide how to go about this, we've recently learnt integration by parts and integration by substitution - I'm not sure what will be easiest to do :')
    If you could talk me through what you're doing that would be extremely helpful - thanks all!!
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    Well it might be easiest to take x as U and (4-3x^2)^-1/2 as dV/dx since then dU/dx will equal 1 which is nice to have

    Integrate dV/dx to get V by letting 4-3x^2 = P so then you have to integrate P^-1/2.dx (remember to get dx in terms of dP so you can integrate)

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    (Original post by eternaforest)
    Well it might be easiest to take x as U and (4-3x^2)^-1/2 as dV/dx since then dU/dx will equal 1 which is nice to have

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    I've tried this but I think I must be integrating (4-3x^2)^1/2 wrong, do you know what v would be equal to?
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    (Original post by VetStudentOf2017)
    I've tried this but I think I must be integrating (4-3x^2)^1/2 wrong, do you know what v would be equal to?
    I edited my post, see if the extra bit helps

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    (Original post by eternaforest)
    I edited my post, see if the extra bit helps

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    That makes sense, thank you I'll give it a try!
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    (Original post by VetStudentOf2017)
    That makes sense, thank you I'll give it a try!
    How did it go?

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    You can do this integral by inspection, or the reverse chain rule. Consider what the derivative of (4-3x^2)^3/2 is and then see how this relates to the integral.
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    Hi guys!
    Sorry it took me so long to update on this, I was having problems accessing the post for some reason

    In the end I integrated by inspection as B_9710 said.

    u = 4-3x^2 and -1/3du = xdx

    so in the end it was -1/6 multiplied by the integral of u^1/2 and I substituted my limits in. I found the correct answer eventually which was 7/9! Thanks for your help
 
 
 
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