I=∫_0 ^( ∞) ((x{(a^2 −b^2 )x−2a^2 x^2 −2b^2 }dx)/((a^2 x^2 +b^2 )^2 {(a^2 −b^2 )x+a^2 x^2 +b^2 }))


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Question Number 141244 by ajfour last updated on 17/May/21
$I=∫_0 ^( ∞) ((x{(a^2 −b^2 )x−2a^2 x^2 −2b^2 }dx)/((a^2 x^2 +b^2 )^2 {(a^2 −b^2 )x+a^2 x^2 +b^2 }))$
$I = \int_{0}^{\infty} \frac{x {(a^{2} - b^{2}) x - 2 a^{2} x^{2} - 2 b^{2}} d x}{{(a^{2} x^{2} + b^{2})}^{2} {(a^{2} - b^{2}) x + a^{2} x^{2} + b^{2}}}$
Commented by ajfour last updated on 17/May/21
this will get me perimeter of ellipse!
	Answered by MJS_new last updated on 17/May/21

	$I_{1} = \int \frac{x}{{(a^{2} x^{2} + b^{2})}^{2}} d x = - \frac{1}{2 a^{2} (a^{2} x^{2} + b^{2})}$ $I_{2} = - \frac{3}{a^{2} - b^{2}} \int \frac{d x}{a^{2} x^{2} + b^{2}} = - \frac{3 \arctan \frac{a x}{b}}{a (a^{2} - b^{2}) b}$ $I_{3} = \frac{3}{a^{2} - b^{2}} \int \frac{d x}{a^{2} x^{2} + (a^{2} - b^{2}) x + b^{2}} =$ $= \frac{3}{(a^{2} - b^{2}) \sqrt{a^{4} - 6 a^{2} b^{2} + b^{4}}} \ln \frac{2 a^{2} x^{2} + a^{2} - b^{2} - \sqrt{a^{4} - 6 a^{2} b^{2} + b^{4}}}{2 a^{2} x^{2} + a^{2} - b^{2} + \sqrt{a^{4} - 6 a^{2} b^{2} + b^{4}}}$ $now calculate within the borders$
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