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Question Number 216754 by Tawa11 last updated on 17/Feb/25
∫_( 0) ^( 1) ((x ln^2 (x))/(1 + x^2 )) dx
$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{x}\:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$
Answered by issac last updated on 18/Feb/25
∫_( 0) ^( 1) ((x ln^2 (x))/(1 + x^2 )) dx u=atan(x) (du/dx)=(1/(x^2 +1)) → du=(dx/(x^2 +1)) ∫_0 ^( (π/4)) tan(u)ln^2 (tan(u))du... so Hard to me.... wolfram alpha′s answer is ∫_0 ^( 1) ((xln^2 (x))/(x^2 +1))dx=(3/(16))𝛇(3) 𝛇(s)=Σ_(h=1) ^∞ (1/h^s )
$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{x}\:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$$${u}=\mathrm{atan}\left({x}\right) \\ $$$$\frac{\mathrm{d}{u}}{\mathrm{d}{x}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:\:\rightarrow\:\mathrm{d}{u}=\frac{\mathrm{d}{x}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\mathrm{tan}\left({u}\right)\mathrm{ln}^{\mathrm{2}} \left(\mathrm{tan}\left({u}\right)\right)\mathrm{d}{u}… \\ $$$$\mathrm{so}\:\mathrm{Hard}\:\mathrm{to}\:\mathrm{me}…. \\ $$$$\mathrm{wolfram}\:\mathrm{alpha}'\mathrm{s}\:\mathrm{answer}\:\mathrm{is} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}\mathrm{ln}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}=\frac{\mathrm{3}}{\mathrm{16}}\boldsymbol{\zeta}\left(\mathrm{3}\right) \\ $$$$\boldsymbol{\zeta}\left({s}\right)=\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}^{{s}} } \\ $$
Commented by Frix last updated on 19/Feb/25
If you cannot solve it, don′t solve it. It′s really that easy. WolframAlpha′s answer is useless, we need the path.
$$\mathrm{If}\:\mathrm{you}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it},\:\mathrm{don}'\mathrm{t}\:\mathrm{solve}\:\mathrm{it}.\:\mathrm{It}'\mathrm{s} \\ $$$$\mathrm{really}\:\mathrm{that}\:\mathrm{easy}.\:\mathrm{WolframAlpha}'\mathrm{s}\:\mathrm{answer} \\ $$$$\mathrm{is}\:\mathrm{useless},\:\mathrm{we}\:\mathrm{need}\:\mathrm{the}\:\mathrm{path}. \\ $$
Commented by mathmax last updated on 23/Feb/25
3/16 ξ(3) is not correct the answer is ln2 −(π/2)+2k_o k_0 is catalan constante
$$\mathrm{3}/\mathrm{16}\:\xi\left(\mathrm{3}\right)\:{is}\:{not}\:{correct}\:{the}\:{answer}\:{is} \\ $$$${ln}\mathrm{2}\:−\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}_{{o}} \:\:\:\:\:{k}_{\mathrm{0}} {is}\:{catalan}\:{constante} \\ $$
Commented by Tawa11 last updated on 23/Feb/25
Please workings sir.
$$\mathrm{Please}\:\mathrm{workings}\:\mathrm{sir}. \\ $$
Answered by Frix last updated on 19/Feb/25
Some steps: 1 ∫((xln^2 x)/(x^2 +1))dx=(1/2)∫((ln^2 x)/(x−i))dx+(1/2)∫((ln^2 x)/(x+i))dx 2 ∫((ln^2 x)/(x+a))dx =^([t=x+a]) ∫((ln^2 (t−a))/t)dt =^([by parts]) =ln (t/a) ln^2 (t−a) −2∫((ln (t/a) ln (t−a))/(t−a))dt 3 ∫((ln (t/a) ln (t−a))/(t−a))dt =^([by parts]) =−ln (t−a) Li_2 (1−(t/a)) +∫((Li_2 (1−(t/a)))/(t−a))dt 4 ∫((Li_2 (1−(t/a)))/(t−a))dt =^([u=1−(t/a)]) ∫((Li_2 u)/u)du=Li_3 u Not sure what you learned and how you′re supposed to get the answer...
$$\mathrm{Some}\:\mathrm{steps}: \\ $$$$\mathrm{1} \\ $$$$\int\frac{{x}\mathrm{ln}^{\mathrm{2}} \:{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{ln}^{\mathrm{2}} \:{x}}{{x}−\mathrm{i}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{ln}^{\mathrm{2}} \:{x}}{{x}+\mathrm{i}}{dx} \\ $$$$\mathrm{2} \\ $$$$\int\frac{\mathrm{ln}^{\mathrm{2}} \:{x}}{{x}+{a}}{dx}\:\overset{\left[{t}={x}+{a}\right]} {=}\:\int\frac{\mathrm{ln}^{\mathrm{2}} \:\left({t}−{a}\right)}{{t}}{dt}\:\overset{\left[\mathrm{by}\:\mathrm{parts}\right]} {=} \\ $$$$=\mathrm{ln}\:\frac{{t}}{{a}}\:\mathrm{ln}^{\mathrm{2}} \:\left({t}−{a}\right)\:−\mathrm{2}\int\frac{\mathrm{ln}\:\frac{{t}}{{a}}\:\mathrm{ln}\:\left({t}−{a}\right)}{{t}−{a}}{dt} \\ $$$$\mathrm{3} \\ $$$$\int\frac{\mathrm{ln}\:\frac{{t}}{{a}}\:\mathrm{ln}\:\left({t}−{a}\right)}{{t}−{a}}{dt}\:\overset{\left[\mathrm{by}\:\mathrm{parts}\right]} {=} \\ $$$$=−\mathrm{ln}\:\left({t}−{a}\right)\:\mathrm{Li}_{\mathrm{2}} \:\left(\mathrm{1}−\frac{{t}}{{a}}\right)\:+\int\frac{\mathrm{Li}_{\mathrm{2}} \:\left(\mathrm{1}−\frac{{t}}{{a}}\right)}{{t}−{a}}{dt} \\ $$$$\mathrm{4} \\ $$$$\int\frac{\mathrm{Li}_{\mathrm{2}} \:\left(\mathrm{1}−\frac{{t}}{{a}}\right)}{{t}−{a}}{dt}\:\overset{\left[{u}=\mathrm{1}−\frac{{t}}{{a}}\right]} {=}\:\int\frac{\mathrm{Li}_{\mathrm{2}} \:{u}}{{u}}{du}=\mathrm{Li}_{\mathrm{3}} \:{u} \\ $$$$\mathrm{Not}\:\mathrm{sure}\:\mathrm{what}\:\mathrm{you}\:\mathrm{learned}\:\mathrm{and}\:\mathrm{how}\:\mathrm{you}'\mathrm{re} \\ $$$$\mathrm{supposed}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{answer}… \\ $$

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