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C3 Integration (again...) watch

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    02∫x/{√(4+x2)}

    This is what I did...

    = ½.02∫2x/(4+x2)}

    = ½02[ln(4+x2)]

    = ½.(ln8 - ln4)

    = ½ln2

    ...but the answer is apparently 2(√2-1)

    Could somebody explain what I've done wrong? Thanks.
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    You forgot about the square-root.

    Try the substitution

    u = 4 + x^2
    du/dx = 2x
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    (Original post by Aired)
    02∫x/{√(4+x2)}

    This is what I did...

    = ½.02∫2x/(4+x2)}

    = ½02[ln(4+x2)]

    = ½.(ln8 - ln4)

    = ½ln2

    ...but the answer is apparently 2(√2-1)

    Could somebody explain what I've done wrong? Thanks.
    Is this solomon A?
    I've just done it.
    It tells you the substitution doesn't it?
    :confused:
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    So you end up with [√u]84

    Which is equal to √8 - √4, and the correct answer.

    Thanks
 
 
 
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