Question Number 47088 by Meritguide1234 last updated on 04/Nov/18
Answered by MJS last updated on 04/Nov/18
![x^2 +ax+b=0 x=−(a/2)±((√(a^2 −4b))/2) ⇒ α=−(a/2)−((√(a^2 −4b))/2); β=−(a/2)+((√(a^2 −4b))/(2()) bx^2 +ax+1=0 x=−(a/(2b))±((√(a^2 −4b))/(2b)); ⇒ γ=(α/b); δ=(β/b) (((x^2 −1)ln x)/((x−α)(x−β)(x−γ)(x−δ)))= =A((ln x)/(x−α))+B((ln x)/(x−β))+C((ln x)/(x−γ))+D((ln x)/(x−δ)) A=((α^2 −1)/((α−β)(α−γ)(α−δ))) B=((β^2 −1)/((β−α)(β−γ)(β−δ))) C=((γ^2 −1)/((γ−α)(γ−β)(γ−δ))) D=((δ^2 −1)/((δ−α)(δ−β)(δ−γ))) ∫((ln x)/(x−λ))dx= [t=x−λ → dx=dt] =∫((ln (t+λ))/t)dt=∫(((ln ((t+λ)/λ))/t)+((ln λ)/t))dt= =∫((ln ((t+λ)/λ))/t)dt+ln λ ∫(dt/t) ln λ ∫(dt/t)=ln λ ln t =ln λ ln ∣x−λ∣ ∫((ln ((t+λ)/λ))/t)dt= [u=−(t/λ) → dt=−λdu] =−∫−((ln (1−u))/u)du= [Dilogarithm] =−Li_2 u =−Li_2 −(t/λ) =−Li_2 ((λ−x)/λ) ∫_0 ^1 ((ln x)/(x−λ))dx=[ln λ ln ∣x−λ∣ −Li_2 ((λ−x)/λ)]_0 ^1 = =(π^2 /6)−ln^2 λ +ln (λ−1) ln λ −Li_2 ((λ−1)/λ) sorry I′m too lazy to finish this, it should be clear now anyway](https://www.tinkutara.com/question/Q47110.png)
$${x}^{\mathrm{2}} +{ax}+{b}=\mathrm{0} \\ $$$${x}=−\frac{{a}}{\mathrm{2}}\pm\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}}\:\Rightarrow\:\alpha=−\frac{{a}}{\mathrm{2}}−\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}};\:\beta=−\frac{{a}}{\mathrm{2}}+\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}\left(\right.} \\ $$$${bx}^{\mathrm{2}} +{ax}+\mathrm{1}=\mathrm{0} \\ $$$${x}=−\frac{{a}}{\mathrm{2}{b}}\pm\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}{b}}}{\mathrm{2}{b}};\:\Rightarrow\:\gamma=\frac{\alpha}{{b}};\:\delta=\frac{\beta}{{b}} \\ $$$$\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{ln}\:{x}}{\left({x}−\alpha\right)\left({x}−\beta\right)\left({x}−\gamma\right)\left({x}−\delta\right)}= \\ $$$$=\mathcal{A}\frac{\mathrm{ln}\:{x}}{{x}−\alpha}+\mathcal{B}\frac{\mathrm{ln}\:{x}}{{x}−\beta}+\mathcal{C}\frac{\mathrm{ln}\:{x}}{{x}−\gamma}+\mathcal{D}\frac{\mathrm{ln}\:{x}}{{x}−\delta} \\ $$$$\mathcal{A}=\frac{\alpha^{\mathrm{2}} −\mathrm{1}}{\left(\alpha−\beta\right)\left(\alpha−\gamma\right)\left(\alpha−\delta\right)} \\ $$$$\mathcal{B}=\frac{\beta^{\mathrm{2}} −\mathrm{1}}{\left(\beta−\alpha\right)\left(\beta−\gamma\right)\left(\beta−\delta\right)} \\ $$$$\mathcal{C}=\frac{\gamma^{\mathrm{2}} −\mathrm{1}}{\left(\gamma−\alpha\right)\left(\gamma−\beta\right)\left(\gamma−\delta\right)} \\ $$$$\mathcal{D}=\frac{\delta^{\mathrm{2}} −\mathrm{1}}{\left(\delta−\alpha\right)\left(\delta−\beta\right)\left(\delta−\gamma\right)} \\ $$$$ \\ $$$$\int\frac{\mathrm{ln}\:{x}}{{x}−\lambda}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}−\lambda\:\rightarrow\:{dx}={dt}\right] \\ $$$$=\int\frac{\mathrm{ln}\:\left({t}+\lambda\right)}{{t}}{dt}=\int\left(\frac{\mathrm{ln}\:\frac{{t}+\lambda}{\lambda}}{{t}}+\frac{\mathrm{ln}\:\lambda}{{t}}\right){dt}= \\ $$$$=\int\frac{\mathrm{ln}\:\frac{{t}+\lambda}{\lambda}}{{t}}{dt}+\mathrm{ln}\:\lambda\:\int\frac{{dt}}{{t}} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{ln}\:\lambda\:\int\frac{{dt}}{{t}}=\mathrm{ln}\:\lambda\:\mathrm{ln}\:{t}\:=\mathrm{ln}\:\lambda\:\mathrm{ln}\:\mid{x}−\lambda\mid \\ $$$$ \\ $$$$\:\:\:\:\:\int\frac{\mathrm{ln}\:\frac{{t}+\lambda}{\lambda}}{{t}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}=−\frac{{t}}{\lambda}\:\rightarrow\:{dt}=−\lambda{du}\right] \\ $$$$\:\:\:\:\:=−\int−\frac{\mathrm{ln}\:\left(\mathrm{1}−{u}\right)}{{u}}{du}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{Dilogarithm}\right] \\ $$$$\:\:\:\:\:=−\mathrm{Li}_{\mathrm{2}} \:{u}\:=−\mathrm{Li}_{\mathrm{2}} \:−\frac{{t}}{\lambda}\:=−\mathrm{Li}_{\mathrm{2}} \:\frac{\lambda−{x}}{\lambda} \\ $$$$ \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:{x}}{{x}−\lambda}{dx}=\left[\mathrm{ln}\:\lambda\:\mathrm{ln}\:\mid{x}−\lambda\mid\:−\mathrm{Li}_{\mathrm{2}} \:\frac{\lambda−{x}}{\lambda}\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\mathrm{ln}^{\mathrm{2}} \:\lambda\:+\mathrm{ln}\:\left(\lambda−\mathrm{1}\right)\:\mathrm{ln}\:\lambda\:−\mathrm{Li}_{\mathrm{2}} \:\frac{\lambda−\mathrm{1}}{\lambda} \\ $$$$ \\ $$$$\mathrm{sorry}\:\mathrm{I}'\mathrm{m}\:\mathrm{too}\:\mathrm{lazy}\:\mathrm{to}\:\mathrm{finish}\:\mathrm{this},\:\mathrm{it}\:\mathrm{should}\:\mathrm{be} \\ $$$$\mathrm{clear}\:\mathrm{now}\:\mathrm{anyway} \\ $$
Commented by Meritguide1234 last updated on 05/Nov/18
