Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 135431 by 777316 last updated on 13/Mar/21

Answered by SEKRET last updated on 13/Mar/21

F(a)= ∫_0 ^( 1) ((ln(ax^2 +1))/((x+1))) dx a=1 F ′(a)= ∫_0 ^1 (x^2 /((x+1)(ax^2 +1))) dx=(1/(a+1))∙∫_0 ^1 ((x/(ax^2 +1))−(1/(ax^2 +1)) + (1/(x+1))) F′(a)= [(1/(a+1))∙((1/(2a))∙ln(ax^2 +1) − (1/( (√a)))∙arctg((√a) ∙x) + ln(x+1))]_0 ^1 F ′(a)= (1/(a+1))∙((1/(2a))∙ln(a+1)−(1/( (√a)))∙arctg((√a)) +ln(2)) F(a)= ∫ ((ln(a+1))/(2a∙(a+1))) da −∫ ((arctg((√a)))/( (√a) ∙(a+1))) da+ln(2)∙∫(1/(a+1))da F(a)= ((−1)/4)∙ln^2 (a+1) − ((L_2 i(−a))/2) − arctg^2 ((√a))+ln(2)∙ln(a+1)+C a=0 F(0)=0 0 = 0−0−0+0+C C=0 F(1)= ((−1)/4)∙ln^2 (2) − (𝛑^2 /(24)) − (𝛑^2 /(16)) + ln^2 (2) F(1) = (3/4)∙ln^2 (2) − ((5𝛑^2 )/(48))

$$\boldsymbol{\mathrm{F}}\left(\boldsymbol{\mathrm{a}}\right)=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{ax}}^{\mathrm{2}} +\mathrm{1}\right)}{\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)}\:\boldsymbol{\mathrm{dx}}\:\:\:\:\:\:\boldsymbol{\mathrm{a}}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{F}}\:'\left(\boldsymbol{\mathrm{a}}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\left(\boldsymbol{\mathrm{ax}}^{\mathrm{2}} +\mathrm{1}\right)}\:\boldsymbol{\mathrm{dx}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}+\mathrm{1}}\centerdot\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{ax}}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{ax}}^{\mathrm{2}} +\mathrm{1}}\:+\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}+\mathrm{1}}\right) \\ $$$$\:\boldsymbol{\mathrm{F}}'\left(\boldsymbol{\mathrm{a}}\right)=\:\left[\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}+\mathrm{1}}\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\mathrm{a}}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{ax}}^{\mathrm{2}} +\mathrm{1}\right)\:−\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{a}}}}\centerdot\boldsymbol{\mathrm{arctg}}\left(\sqrt{\boldsymbol{\mathrm{a}}}\:\centerdot\boldsymbol{\mathrm{x}}\right)\:+\:\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\boldsymbol{\mathrm{F}}\:'\left(\boldsymbol{\mathrm{a}}\right)=\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}+\mathrm{1}}\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{\mathrm{a}}}\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)−\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{a}}}}\centerdot\boldsymbol{\mathrm{arctg}}\left(\sqrt{\boldsymbol{\mathrm{a}}}\right)\:+\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\right) \\ $$$$\:\boldsymbol{\mathrm{F}}\left(\boldsymbol{\mathrm{a}}\right)=\:\int\:\frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)}{\mathrm{2}\boldsymbol{\mathrm{a}}\centerdot\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)}\:\boldsymbol{\mathrm{da}}\:−\int\:\frac{\boldsymbol{\mathrm{arctg}}\left(\sqrt{\boldsymbol{\mathrm{a}}}\right)}{\:\sqrt{\boldsymbol{\mathrm{a}}}\:\centerdot\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)}\:\boldsymbol{\mathrm{da}}+\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\centerdot\int\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}+\mathrm{1}}\boldsymbol{\mathrm{da}} \\ $$$$\boldsymbol{\mathrm{F}}\left(\boldsymbol{\mathrm{a}}\right)=\:\frac{−\mathrm{1}}{\mathrm{4}}\centerdot\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)\:−\:\frac{\boldsymbol{\mathrm{L}}_{\mathrm{2}} \boldsymbol{\mathrm{i}}\left(−\boldsymbol{\mathrm{a}}\right)}{\mathrm{2}}\:−\:\boldsymbol{\mathrm{arctg}}^{\mathrm{2}} \left(\sqrt{\boldsymbol{\mathrm{a}}}\right)+\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\centerdot\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)+\boldsymbol{\mathrm{C}} \\ $$$$\boldsymbol{\mathrm{a}}=\mathrm{0}\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{F}}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\:\mathrm{0}\:=\:\mathrm{0}−\mathrm{0}−\mathrm{0}+\mathrm{0}+\boldsymbol{\mathrm{C}}\:\:\:\:\:\:\boldsymbol{\mathrm{C}}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{F}}\left(\mathrm{1}\right)=\:\frac{−\mathrm{1}}{\mathrm{4}}\centerdot\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{2}\right)\:−\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{24}}\:−\:\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{16}}\:+\:\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\:\boldsymbol{\mathrm{F}}\left(\mathrm{1}\right)\:=\:\frac{\mathrm{3}}{\mathrm{4}}\centerdot\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{2}\right)\:−\:\frac{\mathrm{5}\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{48}} \\ $$

Answered by Ñï= last updated on 13/Mar/21

∫_0 ^1 ((ln(x^2 +1))/(x+1))dx=∫_0 ^1 (1/(x+1))∫_0 ^x ((2y)/(y^2 +1))dydx =∫_0 ^1 (x/(x+1))∫_0 ^1 ((2xy)/(x^2 y^2 +1))dydx =∫_0 ^1 2y∫_0 ^1 (1/(y^2 +1))((1/(x+1))+((x−1)/(x^2 y^2 +1)))dxdy =∫_0 ^1 ((2y)/(y^2 +1))∫_0 ^1 ((1/(x+1))+((x−1)/(x^2 y^2 +1)))dxdy =∫_0 ^1 ((2y)/(y^2 +1)){ln2+(1/(2y^2 ))ln(y^2 +1)−(1/y)tan^(−1) y}dy =(ln2)ln(y^2 +1)∣_0 ^1 +∫_0 ^1 ((ln(y^2 +1))/(y(y^2 +1)))dy−2∫_0 ^1 ((tan^(−1) y)/(y^2 +1))dy =ln^2 2+(1/2)∫_0 ^1 ((ln(y^2 +1))/(y^2 (y^2 +1)))d(y^2 )−(tan^(−1) y)^2 ∣_0 ^1 =ln^2 2+(1/2)∫_0 ^1 ((1/y)−(1/(y+1)))ln(y+1)dy−(π^2 /(16)) =ln^2 2−(1/2)Li_2 (−1)−(1/4)ln^2 (y+1)∣_0 ^1 −(π^2 /(16)) =(3/4)ln^2 2−((5π^2 )/(48))

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}+\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}+\mathrm{1}}\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{2}{y}}{{y}^{\mathrm{2}} +\mathrm{1}}{dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{{x}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{1}}{dydx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{y}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}}\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{1}}\right){dxdy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{y}}{{y}^{\mathrm{2}} +\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{1}}\right){dxdy} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{y}}{{y}^{\mathrm{2}} +\mathrm{1}}\left\{{ln}\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}{y}^{\mathrm{2}} }{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)−\frac{\mathrm{1}}{{y}}\mathrm{tan}^{−\mathrm{1}} {y}\right\}{dy} \\ $$$$=\left({ln}\mathrm{2}\right){ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)}{{y}\left({y}^{\mathrm{2}} +\mathrm{1}\right)}{dy}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tan}^{−\mathrm{1}} {y}}{{y}^{\mathrm{2}} +\mathrm{1}}{dy} \\ $$$$={ln}^{\mathrm{2}} \mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)}{{y}^{\mathrm{2}} \left({y}^{\mathrm{2}} +\mathrm{1}\right)}{d}\left({y}^{\mathrm{2}} \right)−\left(\mathrm{tan}^{−\mathrm{1}} {y}\right)^{\mathrm{2}} \mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$={ln}^{\mathrm{2}} \mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{{y}}−\frac{\mathrm{1}}{{y}+\mathrm{1}}\right){ln}\left({y}+\mathrm{1}\right){dy}−\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$={ln}^{\mathrm{2}} \mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}{Li}_{\mathrm{2}} \left(−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}{ln}^{\mathrm{2}} \left({y}+\mathrm{1}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} −\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{ln}^{\mathrm{2}} \mathrm{2}−\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com