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    Integrate x sqrt (2x-1) dx

    let u=2x-1

    Got the answer as 1/4 [2/5 (2x-1) ^ 5/2 + 2/3 (2x-1) ^ 3/2 +C)

    Taking the fractions outside the brackets..

    1/15 [ (2x-1) ^5/2 + (2x-1) ^3/2 +C]

    Is this correct?
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    Nearly right - the answer itself is right but then the simplification with taking fractions out is wrong.

    Try taking out 2 and a power of 2x-1.
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    (Original post by Ab3rry)
    Nearly right - the answer itself is right but then the simplification with taking fractions out is wrong.

    Try taking out 2 and a power of 2x-1.
    Taking a 2 out from where, you mean from both fractions, 2/5 and 2/3?

    Take a power of 2x-1 outside from the bracket to which it is raised to 5/2?
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    Yeah, take the 2 out from both fractions.

    Taking (2x-1)^5/2 out will give you a negative power in the brackets, which isn't ideal... so try the other option!
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    (Original post by Ab3rry)
    Yeah, take the 2 out from both fractions.

    Taking (2x-1)^5/2 out will give you a negative power in the brackets, which isn't ideal... so try the other option!


    If the 2 is taken out from both fractions, it removes the 1/4 on the front of the eqn right?

    so [ 1/5 (2x+1) ^5/2 + 1/3 (2x+1) ^3/2+C]

    So remove a (2x+1) from (2x+) ^3/2?

    that gives (2x+1) [ 1/5 (2x+1) ^ 5/2 + 1/3 (2x+1) ^1/2 +C] ?

    EDIT: Replace all (2x+1) with (2x-1), whoops my bad :P
 
 
 
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