I believe the answer is 0.5(x(x^2-a)^0.5)-0.5aln(x+(x^2-a)^0.5)

does anyone know how to get this/derive this i.e. not take from tables !!!!

Thanks

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- Dec 2nd 2008, 06:24 AMcoverbandIntegrate Sqrt[x^2-a]?
I believe the answer is 0.5(x(x^2-a)^0.5)-0.5aln(x+(x^2-a)^0.5)

does anyone know how to get this/derive this i.e. not take from tables !!!!

Thanks - Dec 2nd 2008, 07:05 AMJhevon
that is correct.

to get the derivative of the answer, you need to use the product rule and the chain rule. are you familiar with these.

moreover, note that $\displaystyle \frac d{dx} \ln u = \frac {u'}u$ (by the chain rule)

where $\displaystyle u$ is some function of $\displaystyle x$ - Dec 2nd 2008, 07:39 AMChop Suey
- Dec 4th 2008, 12:53 PMMathstud28
try letting $\displaystyle x=\sqrt{a}\cosh(\theta)$

- Dec 4th 2008, 12:56 PMcoverband
mathstud what if a is negative - then we can't consider sqrt(a) ?

- Dec 4th 2008, 12:57 PMMathstud28
- Dec 4th 2008, 12:59 PMcoverband
but we want to integrate Sqrt[x^2-a] and i forsee difficulty for if a<0 we can not use sqrt(a) in our substitutions

- Dec 4th 2008, 01:00 PMMathstud28
- Dec 4th 2008, 01:00 PMcoverband
- Dec 4th 2008, 01:00 PMMathstud28
- Dec 4th 2008, 01:01 PMJameson