Question Number 110964 by mathdave last updated on 01/Sep/20
$${solve}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{ln}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$
Answered by mathdave last updated on 01/Sep/20
![enjoy my solution let x=tany,dx=(1+tan^2 y)dy I=∫_0 ^(π/4) ((tanyln(tany))/((1+tan^2 y)^3 ))×(1+tan^2 y)dy I=∫_0 ^(π/4) ((tan^2 yln(tany))/(sec^4 y))dy=∫_0 ^(π/4) ((tan^2 yln(tany)cos^4 y)/1)dy I=∫_0 ^(π/4) (sinycosy)^2 ln(tany)dy but sinycosy=((sin2y)/2) I=(1/4)∫_0 ^(π/4) sin^2 2yln(tany)dy but cos4y=1−2sin^2 2y,sin^2 2y=((1−cos4y)/2) I=(1/8)∫_0 ^(π/4) (1−cos4y)ln(tany)dy I=(1/8)∫_0 ^(π/4) ln(tany)−(1/8)∫_0 ^(π/4) cos4yln(tany)dy let B=∫_0 ^(π/4) cos4yln(tany)dy using IBP but let ∫dv=∫cos4ydy,v=((sin4y)/4),u=ln(tany),du=((sec^2 y)/(tany))=(1/(sinycosy))=(2/(sin2y)) B=(1/4)[sin4yln(tany)]_0 ^(π/4) −(1/4)∫_0 ^(π/4) ((2sin4y)/(sin2y))dy B=−(1/2)∫_0 ^(π/4) ((sin4y)/(sin2y))dy (let 2y=x,dy=(dx/2)) =−(1/4)∫_0 ^(π/2) ((sin2x)/(sinx))dx=−(1/4)∫_0 ^(π/2) ((2sinxcosx)/(sinx))dx B=−(1/2)∫_0 ^(π/2) cosxdx=−(1/2)[sinx]_0 ^(π/2) =−(1/2) ∵B=∫_0 ^(π/4) cos4yln(tany)dy=−(1/2).........(1) then letA=∫_0 ^(π/4) ln(tany)dy let x=tany,dy=(dx/(1+x^2 )) A=∫_0 ^1 ((lnx)/(1+x^2 ))dx (using series summation) A=∫_0 ^1 (−1)^n Σ_(n=0) ^∞ x^(2n) lnxdx=(−1)^n Σ_(n=0) ^∞ ∫_0 ^1 x^(2n) .x^a dx∣_(a=0) (∂/∂a)∣_(a=0) A(a)=(−1)^n Σ_(n=0) ^∞ (∂/∂a)∫_0 ^1 x^(2n+a) dx (∂/∂a)∣_(a=0) A(a)=(−1)^n (∂/∂a)Σ_(n=0) ^∞ [(1/((2n+a+1)))] A(0)=A=−Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 ))=−G(catalan constant).....(2) but I=(1/8)A−(1/8)B=[−(G/8)−(1/8)(−(1/2))]=[(1/(16))−(1/8)G] ∵∫_0 ^1 ((x^2 lnx)/((1+x^2 )^3 ))dx=[(1/(16))−(1/8)G] solution by mathdave(1/09/2020)](https://www.tinkutara.com/question/Q110974.png)
$${enjoy}\:{my}\:{solution}\: \\ $$$${let}\:{x}=\mathrm{tan}{y},{dx}=\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {y}\right){dy} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{tan}{y}\mathrm{ln}\left(\mathrm{tan}{y}\right)}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {y}\right)^{\mathrm{3}} }×\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {y}\right){dy} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{tan}^{\mathrm{2}} {y}\mathrm{ln}\left(\mathrm{tan}{y}\right)}{\mathrm{sec}^{\mathrm{4}} {y}}{dy}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{tan}^{\mathrm{2}} {y}\mathrm{ln}\left(\mathrm{tan}{y}\right)\mathrm{cos}^{\mathrm{4}} {y}}{\mathrm{1}}{dy} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{sin}{y}\mathrm{cos}{y}\right)^{\mathrm{2}} \mathrm{ln}\left(\mathrm{tan}{y}\right){dy} \\ $$$${but}\:\mathrm{sin}{y}\mathrm{cos}{y}=\frac{\mathrm{sin2}{y}}{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{sin}^{\mathrm{2}} \mathrm{2}{y}\mathrm{ln}\left(\mathrm{tan}{y}\right){dy} \\ $$$${but}\:\mathrm{cos4}{y}=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{2}{y},\mathrm{sin}^{\mathrm{2}} \mathrm{2}{y}=\frac{\mathrm{1}−\mathrm{cos4}{y}}{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}−\mathrm{cos4}{y}\right)\mathrm{ln}\left(\mathrm{tan}{y}\right){dy} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tan}{y}\right)−\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{cos4}{y}\mathrm{ln}\left(\mathrm{tan}{y}\right){dy} \\ $$$${let}\:\:{B}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{cos4}{y}\mathrm{ln}\left(\mathrm{tan}{y}\right){dy} \\ $$$${using}\:{IBP}\:\:\:\:{but}\:{let}\: \\ $$$$\int{dv}=\int\mathrm{cos4}{ydy},{v}=\frac{\mathrm{sin4}{y}}{\mathrm{4}},{u}=\mathrm{ln}\left(\mathrm{tan}{y}\right),{du}=\frac{\mathrm{sec}^{\mathrm{2}} {y}}{\mathrm{tan}{y}}=\frac{\mathrm{1}}{\mathrm{sin}{y}\mathrm{cos}{y}}=\frac{\mathrm{2}}{\mathrm{sin2}{y}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{sin4}{y}\mathrm{ln}\left(\mathrm{tan}{y}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{2sin4}{y}}{\mathrm{sin2}{y}}{dy} \\ $$$${B}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{sin4}{y}}{\mathrm{sin2}{y}}{dy}\:\:\:\:\left({let}\:\mathrm{2}{y}={x},{dy}=\frac{{dx}}{\mathrm{2}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin2}{x}}{\mathrm{sin}{x}}{dx}=−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2sin}{x}\mathrm{cos}{x}}{\mathrm{sin}{x}}{dx} \\ $$$${B}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}{xdx}=−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\because{B}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{cos4}{y}\mathrm{ln}\left(\mathrm{tan}{y}\right){dy}=−\frac{\mathrm{1}}{\mathrm{2}}………\left(\mathrm{1}\right) \\ $$$${then}\:{letA}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tan}{y}\right){dy} \\ $$$${let}\:{x}=\mathrm{tan}{y},{dy}=\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:\:\left({using}\:{series}\:{summation}\right) \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} \mathrm{ln}{xdx}=\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}} .{x}^{{a}} {dx}\mid_{{a}=\mathrm{0}} \\ $$$$\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} {A}\left({a}\right)=\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\partial}{\partial{a}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+{a}} {dx} \\ $$$$\frac{\partial}{\partial{a}}\mid_{{a}=\mathrm{0}} {A}\left({a}\right)=\left(−\mathrm{1}\right)^{{n}} \frac{\partial}{\partial{a}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\left(\mathrm{2}{n}+{a}+\mathrm{1}\right)}\right] \\ $$$${A}\left(\mathrm{0}\right)={A}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=−{G}\left({catalan}\:{constant}\right)…..\left(\mathrm{2}\right) \\ $$$${but}\:{I}=\frac{\mathrm{1}}{\mathrm{8}}{A}−\frac{\mathrm{1}}{\mathrm{8}}{B}=\left[−\frac{{G}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\right]=\left[\frac{\mathrm{1}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{8}}{G}\right] \\ $$$$\because\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \mathrm{ln}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx}=\left[\frac{\mathrm{1}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{8}}{G}\right] \\ $$$${solution}\:{by}\:{mathdave}\left(\mathrm{1}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 01/Sep/20
$${okay}.. \\ $$