please ∫_0 ^∞ ((xlogx)/((1+x^2 )^2 ))dx=


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Question Number 11420 by ainstain last updated on 25/Mar/17

$p l e a s e$ $\int_{0}^{\infty} \frac{x l o g x}{{(1 + x^{2})}^{2}} d x =$
Commented by FilupS last updated on 26/Mar/17

$is \log (x) in base 10 or base e ?$
	Answered by sm3l2996 last updated on 25/Mar/17
	${_(v′=(x/((1+x^2 )^2 ))) ^(u=log(x)) ⇒{_(v=((−1)/(2(1+x^2 )))) ^(u^′ =(1/x)) I=∫_0 ^∞ ((xlog(x))/((1+x^2 )^2 ))dx=[((−log(x))/(2(1+x^2 )))]_0 ^∞ +∫_0 ^∞ (dx/(2x(1+x^2 ))) I=lim_(x→∞) (−((log(x))/(2(1+x^2 ))))−lim_(x→0) (−((log(x))/(2(1+x^2 ))))+(1/2)∫_0 ^∞ (dx/(x(1+x^2 ))) (1/(x(1+x^2 )))=(a/x)+((bx+c)/(1+x^2 )) a=1; {_(((c−b)/2)−1=((−1)/2)) ^(((b+c)/2)+1=(1/2)) ⇔{_(c−b=1) ^(b+c=−1) b=−1; c=0 so: (1/(x(1+x^2 )))=(1/x)−(x/(1+x^2 )) ∫_0 ^∞ (dx/(x(1+x^2 )))=∫_0 ^∞ (dx/x)−∫_0 ^∞ ((xdx)/(1+x^2 )) =[log∣x∣−(1/2)log∣1+x^2 ∣]_0 ^∞ =lim_(x→+∞) log((x/(√(1+x^2 ))))−lim_(x→0) log((x/(√(1+x^2 )))) I=lim_(x→+∞) log((x/(√(1+x^2 ))))−lim_(x→0) (log((x/(√(1+x^2 ))))+((log(x))/(2(1+x^2 ))))$
	${_{v^{'} = \frac{x}{{(1 + x^{2})}^{2}}}^{u = \log (x)} \Rightarrow {_{v = \frac{- 1}{2 (1 + x^{2})}}^{u^{^{'}} = \frac{1}{x}}$ $I = \int_{0}^{\infty} \frac{xlog (x)}{{(1 + x^{2})}^{2}} dx = {[\frac{- \log (x)}{2 (1 + x^{2})}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{dx}{2 x (1 + x^{2})}$ $I = \lim_{x \to \infty} (- \frac{\log (x)}{2 (1 + x^{2})}) - \lim_{x \to 0} (- \frac{\log (x)}{2 (1 + x^{2})}) + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{x (1 + x^{2})}$ $\frac{1}{x (1 + x^{2})} = \frac{a}{x} + \frac{bx + c}{1 + x^{2}}$ $a = 1;$ ${_{\frac{c - b}{2} - 1 = \frac{- 1}{2}}^{\frac{b + c}{2} + 1 = \frac{1}{2}} \Leftrightarrow {_{c - b = 1}^{b + c = - 1}$ $b = - 1; c = 0$ $so : \frac{1}{x (1 + x^{2})} = \frac{1}{x} - \frac{x}{1 + x^{2}}$ $\int_{0}^{\infty} \frac{dx}{x (1 + x^{2})} = \int_{0}^{\infty} \frac{dx}{x} - \int_{0}^{\infty} \frac{xdx}{1 + x^{2}}$ $= {[\log ∣ x ∣ - \frac{1}{2} \log ∣ 1 + x^{2} ∣]}_{0}^{\infty}$ $= \underset{x \to + \infty}{\lim l o g} (\frac{x}{\sqrt{1 + x^{2}}}) - \underset{x \to 0}{\lim l o g} (\frac{x}{\sqrt{1 + x^{2}}})$ $I = \underset{x \to + \infty}{\lim l o g} (\frac{x}{\sqrt{1 + x^{2}}}) - \lim_{x \to 0} (\log (\frac{x}{\sqrt{1 + x^{2}}}) + \frac{\log (x)}{2 (1 + x^{2})})$
	Commented by sm3l2996 last updated on 25/Mar/17

	$working$
	Answered by ajfour last updated on 26/Mar/17

	$zero$
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