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Integreation Question

I'm as good as stuck with this one.
8π∫0∞t3(4+t2)52dt\text 8\pi\int_{0}^{\infty}\frac{t^3}{(4+t^2)^\frac{5}{2 }} dt
Reply 1
Avatar for BCHL85
BCHL85
12
Are you really 11? If so you are genious eventhough you can't do this question
-----
Unparseable latex formula:

t^3 = t(t^2 + 4) - 4t\\[br]So\ \frac{t^3}{52(t^2 + 4)} = \frac{t}{52} - \frac{4t}{52(t^2+4)


I guess you can do it from now, as I believed you are genious :biggrin:
Reply 2
Avatar for Dimez
Dimez
2
BCHL85
Are you really 11? If so you are genious eventhough you can't do this question
-----
Unparseable latex formula:

t^3 = t(t^2 + 4) - 4t\\[br]So\ \frac{t^3}{52(t^2 + 4)} = \frac{t}{52} - \frac{4t}{52(t^2+4)


I guess you can do it from now, as I believed you are genious :biggrin:


*Genius

He's bluffing, his actual age is 16.
Reply 3
Avatar for BCHL85
BCHL85
12
Aha, now we have another genious here, 1 year old that can read + type
lol.
He's bluffing or you are?
Reply 4
Avatar for Dimez
Dimez
2
BCHL85
Aha, now we have another genious here, 1 year old that can read + type
lol.
He's bluffing or you are?


*Genius

Lol, I believe we both are. Or perhaps I'm one precocious little child? Who knows?
Reply 5
Avatar for BCHL85
BCHL85
12
sorry, keeping my mistake on genius :p:
well, at least you are genius in bluffing
lol
Reply 6
Avatar for Dimez
Dimez
2
BCHL85
sorry, keeping my mistake on genius :p:
well, at least you are genius in bluffing
lol


I concur. :wink:

You're probably better than me at Mathematics, consider that an accolade. :biggrin:
Reply 7
Avatar for Bleh
Bleh
[br][br]8π∫0∞t3(4+t2)5/2dt=[br]8π2∫4∞x−4x5/2dx=[br]4π∫4∞x−5/2(x−4)dx=[br]4π∫4∞(x−3/2−4x−5/2)dx=[br]4π(8/3x−5/2−2x−1/2)4∞=[br]4π(−(1/3−1))=[br]8π3[br][br][br] [br]8\pi\int_0^{\infty}\frac{t^3}{(4+t^2)^{5/2}}dt =[br]\frac{8\pi}{2}\int_4^{\infty}\frac{x-4}{x^{5/2}}dx \\ =[br]4\pi\int_4^{\infty}x^{-5/2}(x-4)dx \\ =[br]4\pi\int_4^{\infty}(x^{-3/2}-4x^{-5/2})dx \\ =[br]4\pi\left(8/3x^{-5/2}-2x^{-1/2}\right)_4^{\infty} \\ =[br]4\pi\left(-(1/3-1)\right) \\ =[br]\frac{8\pi}{3}[br] [br]

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