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C3 Integration (again...)

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.

Reply 1

Jonny W

You forgot about the square-root.

Try the substitution

u = 4 + x^2
du/dx = 2x

Reply 2

Christophicus

Aired

Is this solomon A?
I've just done it.
It tells you the substitution doesn't it?
:confused:

Reply 3

Aired

So you end up with [√u]⁸₄

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

Thanks

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