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Question Number 115438 by A8;15: last updated on 25/Sep/20

	Answered by Olaf last updated on 26/Sep/20

	$t h e f u n c t i o n i s o d d$ $a n d t h e d o m a i n i s s y m m e t r i c a l$ $s o t h e r e s u l t i s 0 .$ $Sorry I^{'} m wrong .$ $see mister abdo .$
	Commented by mathmax by abdo last updated on 26/Sep/20
	$here {x^3 } mean x^3 −[x^3 ]...!$
	$here {x^{3}} mean x^{3} - [x^{3}] . . .!$
	Commented by Olaf last updated on 26/Sep/20

	$sorry mister!$ $I^{'} m french . In my country we {don}^{'} t$ $use the same notations . Sorry!$
	Commented by Bird last updated on 26/Sep/20

	$n e v e r m i n d$
	Answered by mathmax by abdo last updated on 26/Sep/20
	$I =∫_(−1) ^1 (({x^3 }(x^4 +1))/(x^6 +1))dx ⇒I =∫_(−1) ^1 (((x^3 −[x^3 ](x^4 +1))/(x^6 +1))dx =∫_(−1) ^(1 ) ((x^3 (x^4 +1))/(x^6 +1))dx(→=0 odd function) −∫_(−1) ^1 (([x^3 ](x^4 +1))/(x^6 +1))dx we haveA= ∫_(−1) ^1 (([x^3 ](x^4 +1))/(x^6 +1))dx =_(x=−t) ∫_(−1) ^1 (([−t^3 ](t^4 +1))/(t^6 +1))dt ⇒ 2A =∫_(−1) ^1 ((([x^3 ]+[−x^3 ])(x^4 +1))/(x^6 +1))dx =∫_(−1) ^0 ((([x^3 ]+[−x^3 ])(x^4 +1))/(x^6 +1))dx(x=−t) +∫_0 ^1 ((([x^3 ]+[−x^3 ])(x^4 +1))/(x^6 +1))dx =2∫_0 ^1 ((([x^3 ]+[−x^3 ])(x^4 +1))/(x^6 +1))dx =−2 ∫_0 ^1 ((1+x^4 )/(1+x^6 )) dx =−2 ∫_0 ^1 (1+x^4 )Σ_(n=0) ^∞ (−1)^n x^(6n) =−2 Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 (x^(6n) +x^(6n+4) )dx =−2 Σ_(n=0) ^∞ (((−1)^n )/(6n+1))−2 Σ_(n=0) ^∞ (((−1)^n )/(6n+5)) ...be continued we can calculate ∫_0 ^1 ((1+x^4 )/(1+x^6 ))dx by elementary functions...$
	$I = \int_{- 1}^{1} \frac{{x^{3}} (x^{4} + 1)}{x^{6} + 1} dx \Rightarrow I = \int_{- 1}^{1} \frac{(x^{3} - [x^{3}] (x^{4} + 1)}{x^{6} + 1} dx$ $= \int_{- 1}^{1} \frac{x^{3} (x^{4} + 1)}{x^{6} + 1} dx (\to= 0 odd function) - \int_{- 1}^{1} \frac{[x^{3}] (x^{4} + 1)}{x^{6} + 1} dx$ $we haveA = \int_{- 1}^{1} \frac{[x^{3}] (x^{4} + 1)}{x^{6} + 1} dx =_{x = - t} \int_{- 1}^{1} \frac{[- t^{3}] (t^{4} + 1)}{t^{6} + 1} dt \Rightarrow$ $2 A = \int_{- 1}^{1} \frac{([x^{3}] + [- x^{3}]) (x^{4} + 1)}{x^{6} + 1} dx = \int_{- 1}^{0} \frac{([x^{3}] + [- x^{3}]) (x^{4} + 1)}{x^{6} + 1} dx (x = - t)$ $+ \int_{0}^{1} \frac{([x^{3}] + [- x^{3}]) (x^{4} + 1)}{x^{6} + 1} dx$ $= 2 \int_{0}^{1} \frac{([x^{3}] + [- x^{3}]) (x^{4} + 1)}{x^{6} + 1} dx = - 2 \int_{0}^{1} \frac{1 + x^{4}}{1 + x^{6}} dx$ $= - 2 \int_{0}^{1} (1 + x^{4}) \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{6 n} = - 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} (x^{6 n} + x^{6 n + 4}) dx$ $= - 2 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{6 n + 1} - 2 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{6 n + 5} . . . be continued$ $we can calculate \int_{0}^{1} \frac{1 + x^{4}}{1 + x^{6}} dx by elementary functions . . .$
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