Question

(16-9x^2) ^(3/2)/ x^6

Answer 1

please help

Answer 2

my only though it in the numerator

\[\frac{(16 -9x^2)^{3/2}}{x^6} = \frac{(4-3x)^{3/2}(4 + 3x)^{3/2}}{x^6}\]

Answer 3

or as an alternative

\[\frac{\sqrt{(16 - 9x^2)^3}}{x^6}= \frac{(16-9x^2)\sqrt{(16-9x^2}}{x^6}\]

Answer 4

whats next then?

Answer 5

ok... this can be a split fraction

\[\frac{16\sqrt{16 - 9x^2}}{x^6} - \frac{9x^2\sqrt{16 - 9x^2}}{x^6}\]

eliminate the common factor of x^2 in the 2nd term leaves

\[\frac{16\sqrt{16 - 9x^2}}{x^6} - \frac{9\sqrt{16-9x^2}}{x^4}\]

I think the original form is a lot friendlier

Answer 6

http://www.wolframalpha.com/input/?i=integral+of+%2816-9x%5E2%29+%5E%283%2F2%29%2F+x%5E6


please do check this solution is this right?

Answer 7

lol... just trust wolframalpha

Answer 8

ok