anti-deriv of 1/ (x sqrt(x^2 - 16) dx?

Question

but how would the anti-deriv of it look midway before you use arctan? how would it look in the du and u form i mean.

Answer 1

u^2 = x^2-16
x^2 = u^2+16
u du = x dx

∫1/ (x sqrt(x^2 - 16) dx
=∫1/ (x^2 sqrt(x^2 - 16) xdx
=∫1/(u^2+16) du
=(1/4) arctan(u/4) + c
=(1/4) arctan(sqrt(x^2 - 16)/4) + c

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check:
[(1/4) arctan(sqrt(x^2 - 16)/4)]'
= (1/4)1/[1+(x^2-16)/16][(1/4)x/sqrt(x^2-16)]
=1/[xsqrt(x^2-16)]

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awesomedu,
How did you get your answer? Do you know your answer is exactly the same as mine? Can you prove it?

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Mr.Scient,
Great idea to use arcsec. But your answer is not right.
1/[x sqrt(x^2 - 16)]
=1/[4x sqrt(x^2/4^2 - 1)]
=1/[16(x/4)sqrt(x/4)^2 - 1)]

Therefore,
∫1/ (x sqrt(x^2 - 16) dx
=(1/16)arcsec(x/4) + c

Answer 2

use arctan law, you should get:

(-1/4)arctan(4/sqrt(x^2 - 16)) + c

Answer 3

(1/4)*arcsec(x/4) + C

Answer 4

I'm completely sick of this question,this must be the 20th time that it has been asked.It is just so boring to repeat.Just look at my last 19 answers!!

Answer 5

sahsjing is correct. Nice substitution. That was the key.