Question

Integral of 1/x^3sqrt(x^2-16)

Answer 1

I dont know where to start. Does this use trigonometric substitution?

Answer 2

\[\int\limits \frac{ 1 }{ x ^{3}\sqrt{x ^{2}-16} }dx\]

Answer 3

i was curious how well a decomp would work

Answer 4

what trug sub did you have in mind?

Answer 5

Whats throwing me off is the x^3 out in front of the radical. Ive never seen a trigonometric substitution problem that has that so I dont know how to handle it.

Answer 6

well, it does come from an inverse trig construction

Answer 7

You could do x = 4sec(theta).

Answer 8

sounds plausible

Answer 9

\[\int\limits \frac{ 1 }{ x ^{3}\sqrt{x ^{2}-16} }dx\]
\[\int\limits \frac{ 1 }{ 4^2sec^{3}(t)\sqrt{16sec^2(t)-16} }dx\]
\[dx=4sec(t)tan(t)~dt\]
\[\int\limits \frac{ 4sec(t)tan(t)~dt }{ 4^2sec^{3}(t)\sqrt{16sec^2(t)-16} }\]
\[\int\limits \frac{ 4sec(t)tan(t)~dt }{4^2 sec^{3}(t)~4tan^2(t) }\]
\[\int\limits \frac{ dt }{4^2 sec^{2}(t)~tan(t) }\]
\[\frac1{16}\int\limits cos^3(t)csc(t)dt \]
hmmm

Answer 10

4^3, but yeah, that may not be the most oportune trig sub

Answer 11

\[u = \sqrt{x^2-16}~:~\frac{du}{x} =\frac{dx}{\sqrt{x^2-16}} \]
\[\frac{du}{x^4u}\]
\[u^2 - 16 = x^2~:~x^4=u^4-32u^2+16^2\]
am i missing something in the process, i cant seem to get the (u^2-16)

Answer 12

+16 on the left, and +32 on the right but still

Answer 13

unless i messed up losers thought, i cant think of a simple way to address it

Answer 14

how did you get 4tan^2 from the radical?