Find: 𝛀 =∫_( 0) ^( ∞) ((x arctan(x))/((x + 1)(x^2 + 1))) dx


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Question Number 159292 by HongKing last updated on 15/Nov/21

$Find : Ω = \underset{0}{\int^{\infty}} \frac{x \arctan (x)}{(x + 1) (x^{2} + 1)} dx$
	Answered by mindispower last updated on 15/Nov/21
	$=∫_0 ^∞ ((arctan(x))/(1+x^2 ))−((arctan(x))/((1+x)(1+x^2 )))dx =(π^2 /8)−∫_0 ^(π/2) ((xcos(x))/(cos(x)+sin(x)))dx∣_(=A) B=∫_0 ^(π/2) ((xsin(x))/(sin(x)+cos(x))) A+B=(π^2 /8) A−B=∫_0 ^(π/2) ((x(cos(x)−sin(x)))/(cos(x)+sin(x)))dx =[xln(cos(x)+sin(x))]_0 ^(π/2) −∫_0 ^(π/2) ln(cos(x)+sin(x){dx =−∫_0 ^(π/2) ln((√2)sin(x+(π/4)))d=−(π/2)ln((√2))−∫_(π/4) ^((3π)/4) ln(sin(x))dx =−∫_(π/4) ^(π/2) ln(sin(x))dx−∫_0 ^(π/4) lncos(x) =−2∫_0 ^(π/4) ln(cos(x))dx=−2.(1/4)(2G−πln(2)) catalan constant A=(1/2)(A−B+A+B)=(1/2)((π^2 /8)−(π/2)ln((√2))−G+(π/2)ln(2)) =(π^2 /(16))+(π/8)ln(2)−(G/2) Ω=(π^2 /(16))+(G/2)−((πln(2))/8)$
	$= \int_{0}^{\infty} \frac{a r c t a n (x)}{1 + x^{2}} - \frac{a r c t a n (x)}{(1 + x) (1 + x^{2})} d x$ $= \frac{π^{2}}{8} - \int_{0}^{\frac{π}{2}} \frac{x c o s (x)}{c o s (x) + s i n (x)} d x ∣_{= A}$ $B = \int_{0}^{\frac{π}{2}} \frac{x s i n (x)}{s i n (x) + c o s (x)}$ $A + B = \frac{π^{2}}{8}$ $A - B = \int_{0}^{\frac{π}{2}} \frac{x (c o s (x) - s i n (x))}{c o s (x) + s i n (x)} d x$ $= {[x l n (c o s (x) + s i n (x))]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} l n (c o s (x) + s i n (x) {d x$ $= - \int_{0}^{\frac{π}{2}} l n (\sqrt{2} s i n (x + \frac{π}{4})) d = - \frac{π}{2} l n (\sqrt{2}) - \int_{\frac{π}{4}}^{\frac{3 π}{4}} l n (s i n (x)) d x$ $= - \int_{\frac{π}{4}}^{\frac{π}{2}} l n (s i n (x)) d x - \int_{0}^{\frac{π}{4}} l n c o s (x)$ $= - 2 \int_{0}^{\frac{π}{4}} l n (c o s (x)) d x = - 2 . \frac{1}{4} (2 G - π l n (2))$ $c a t a l a n c o n s t a n t$ $A = \frac{1}{2} (A - B + A + B) = \frac{1}{2} (\frac{π^{2}}{8} - \frac{π}{2} l n (\sqrt{2}) - G + \frac{π}{2} l n (2))$ $= \frac{π^{2}}{16} + \frac{π}{8} l n (2) - \frac{G}{2}$ $Ω = \frac{π^{2}}{16} + \frac{G}{2} - \frac{π l n (2)}{8}$
	Commented by HongKing last updated on 15/Nov/21

	$cool my dear Ser thank you so much$
	Commented by mindispower last updated on 15/Nov/21

	$y o u a r e w e l c o m$ $h a v e a n i c e d a y$
	Commented by HongKing last updated on 15/Nov/21

	$thank you very much my dear S er$
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