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    guys it lloks simple but i been getting different answers to these intrgration problems! pl help

    show that ∫(x²)÷(√(x^6-1)) = 1/3 arcosh x³+ c

    ∫(4x)÷(√(x²+9)) = 8 + In3

    thanks
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    For the first, let x³= u
    3x² dx = du

    = 1/31/√(u² - 1)
    = 1/3arcosh u + c
    = 1/3 arcosh x³+c

    For the second...I get
    u= x²
    du = 2x dx

    = ∫2/√(u + 9)

    v= u+9
    du = dv
    = ∫2/√(v)
    = 4√v
    =4√(x² + 9)

    Cant see how x was supposed to be eliminated in your answer...
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    thanks 4 ur help. the limits in the 2nd problem are 4 to 0.
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    done it thanks
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    (Original post by chubby)
    done it thanks
    How come? As your question was wrong?
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    yeah i copied the q wrongly on to TSR. The numerator was 1+4x not 4x :p:
 
 
 

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