0-log-x-1-x-2-2-dx-

Question Number 150404 by mathdanisur last updated on 12/Aug/21

Ω = \underset{0}{\int^{\infty}} \frac{\log (x)}{{(1 + x^{2})}^{2}} dx = ?

Answered by Ar Brandon last updated on 12/Aug/21

Ω (a) = \int_{0}^{\infty} \frac{\ln x}{a^{2} + x^{2}} d x, x = a \tan ϑ

= \int_{0}^{\frac{π}{2}} \frac{\ln (a \tan ϑ)}{a^{2} + a^{2} \tan^{2} ϑ} \cdot a \sec^{2} ϑ d ϑ

= \frac{1}{a} \int_{0}^{\frac{π}{2}} (\ln a + \ln (\tan ϑ)) d ϑ = \frac{π \ln a}{2 a}

\Rightarrow Ω^{'} (a) = \frac{π}{2} (\frac{1}{a^{2}} - \frac{\ln a}{a^{2}}) = - \int_{0}^{\infty} \frac{2 a \ln x}{{(a^{2} + x^{2})}^{2}} d x

\Rightarrow \int_{0}^{\infty} \frac{\ln x}{{(a^{2} + x^{2})}^{2}} d x = - \frac{π}{4 a} (\frac{1}{a^{2}} - \frac{\ln a}{a^{2}})

\Rightarrow \int_{0}^{\infty} \frac{\ln x}{{(1 + x^{2})}^{2}} d x = - \frac{π}{4}

Commented by mathdanisur last updated on 12/Aug/21

Thank You Ser

Answered by ajfour last updated on 12/Aug/21

Ω = \ln x \int \frac{d x}{{(1 + x^{2})}^{2}}

+ \int {\frac{1}{x} \int \frac{d x}{{(1 + x^{2})}^{2}}} d x

Ω = G \ln x + \int \frac{G d x}{x}

G = \int \frac{d x}{{(1 + x^{2})}^{2}} = \frac{1}{2 x} (- \frac{1}{1 + x^{2}})

- \frac{1}{2} \int \frac{d x}{x^{2} (1 + x^{2})}

l e t x = \tan θ

G = \frac{1}{2} \int (1 + \cos 2 θ) d θ

= \frac{2 θ + \sin 2 θ}{4} = \frac{\tan^{- 1} x}{2} + \frac{x}{2 (1 + x^{2})}

2 (G \ln x) = (\ln x) \tan^{- 1} x

+ \frac{x \ln x}{1 + x^{2}}

2 \int \frac{G d x}{x} = \int {\frac{\tan^{- 1} x}{x} + \frac{1}{1 + x^{2}}} d x

= H + \tan^{- 1} x + c

H = \ln x \tan^{- 1} x - \int \frac{\ln x d x}{(1 + x^{2})}

Ω = \frac{\ln x}{2 x} (- \frac{1}{1 + x^{2}}) - \frac{\ln x}{2} \int \frac{d x}{x^{2} (1 + x^{2})}

+ \frac{\tan^{- 1} x}{2} + \frac{1}{2} \ln x \tan^{- 1} x

- \frac{1}{2} \int \frac{\ln x d x}{(1 + x^{2})} + c

\dots

Commented by mathdanisur last updated on 12/Aug/21

Thankhou Ser, ans . ?

Answered by mnjuly1970 last updated on 12/Aug/21

- \frac{π}{4} \dots .

Answered by Lordose last updated on 12/Aug/21

Ω = \int_{0}^{\infty} \frac{\log (x)}{{(1 + x^{2})}^{2}} dx \overset{x = \tan θ}{=} \int_{0}^{\frac{π}{2}} \frac{\log (\tan θ)}{\sec^{2} θ}

Ω = \int_{0}^{\frac{π}{2}} \cos^{2} θ \log (\sin θ) d θ - \int_{0}^{\frac{π}{2}} \cos^{2} θ \log (\cos θ) d θ

Ω = \frac{\partial}{\partial a} \int_{0}^{\frac{π}{2}} \cos^{2} θ \sin^{a} (θ) d θ ∣_{a = 0} - \frac{\partial}{\partial a} \int_{0}^{\frac{π}{2}} \cos^{a} (θ) d θ ∣_{a = 2}

B (x, y) = 2 \int_{0}^{\frac{π}{2}} \sin^{2 x - 1} (t) \cos^{2 y - 1} (t) dt

Ω = \frac{\partial}{\partial a} (\frac{1}{2} B (\frac{a + 1}{2}, \frac{3}{2})) ∣_{a = 0} - \frac{\partial}{\partial a} (\frac{1}{2} B (\frac{1}{2}, \frac{a + 1}{2})) ∣_{a = 2}

Ω = - \frac{π}{4}

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