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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

x2+ax+b=0x=−a2±a2−4b2⇒α=−a2−a2−4b2;β=−a2+a2−4b2(bx2+ax+1=0x=−a2b±a2−4b2b;⇒γ=αb;δ=βb(x2−1)lnx(x−α)(x−β)(x−γ)(x−δ)==Alnxx−α+Blnxx−β+Clnxx−γ+Dlnxx−δA=α2−1(α−β)(α−γ)(α−δ)B=β2−1(β−α)(β−γ)(β−δ)C=γ2−1(γ−α)(γ−β)(γ−δ)D=δ2−1(δ−α)(δ−β)(δ−γ)∫lnxx−λdx=[t=x−λ→dx=dt]=∫ln(t+λ)tdt=∫(lnt+λλt+lnλt)dt==∫lnt+λλtdt+lnλ∫dttlnλ∫dtt=lnλlnt=lnλln∣x−λ∣∫lnt+λλtdt=[u=−tλ→dt=−λdu]=−∫−ln(1−u)udu=[Dilogarithm]=−Li2u=−Li2−tλ=−Li2λ−xλ∫10lnxx−λdx=[lnλln∣x−λ∣−Li2λ−xλ]01==π26−ln2λ+ln(λ−1)lnλ−Li2λ−1λsorryI′mtoolazytofinishthis,itshouldbeclearnowanyway

Commented by Meritguide1234 last updated on 05/Nov/18

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