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# C3 Integration (again...) watch

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1. 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.
2. You forgot about the square-root.

Try the substitution

u = 4 + x^2
du/dx = 2x
3. (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?
4. So you end up with [√u]84

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

Thanks

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Updated: December 21, 2005
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