∫((x^2 dx)/( (√(x^2 −16)))) = ?


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Question Number 199903 by frankpenredpen last updated on 11/Nov/23

$\int \frac{x^{2} d x}{\sqrt{x^{2} - 16}} = ?$
Commented by cortano12 last updated on 11/Nov/23

$why \partial x ?$
	Answered by frankpenredpen last updated on 11/Nov/23

	$w i t h o u t i n t e g r a t i o n f o r p a r t s$
	Answered by Frix last updated on 11/Nov/23

	$\int \frac{x^{2}}{\sqrt{x^{2} - 16}} d x \overset{t = x + \sqrt{x^{2} - 16}}{=}$ $= \int (\frac{t}{4} + \frac{64}{t^{3}} + \frac{8}{t}) d t = \frac{t^{2}}{8} - \frac{32}{t^{2}} + 8 \ln t =$ $= \frac{x \sqrt{x^{2} - 16}}{2} + 8 \ln (x + \sqrt{x^{2} - 16}) + C$
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