Question Number 210181 by 073 last updated on 02/Aug/24
Commented by Frix last updated on 02/Aug/24
$$\mathrm{I}\:\mathrm{get}\:\frac{\pi\mathrm{ln}\:\mathrm{2}}{\mathrm{8}} \\ $$
Commented by 073 last updated on 02/Aug/24
$${solution}??? \\ $$
Commented by Frix last updated on 03/Aug/24
![f(α)=∫_0 ^1 ((ln (αx+1))/(x^2 +1))dx f(0)=0 f ′(α)=(d/dα)[∫_0 ^1 ((ln (αx+1))/(x^2 +1))dx]= =∫_0 ^1 (∂/∂α)[((ln (αx+1))/(x^2 +1))]dx= =∫_0 ^1 (x/((αx+1)(x^2 +1)))dx= =(1/(α^2 +1))∫_0 ^1 ((x+α)/(x^2 +1))dx−(α/(α^2 +1))∫_0 ^1 (dx/(αx+1))= =[((ln (x^2 +1))/(2(α^2 +1)))+((αtan^(−1) x)/(α^2 +1))−((ln (αx+1))/(α^2 +1))]_0 ^1 = =((απ+2ln 2)/(4(α^2 +1)))−((ln (α+1))/(α^2 +1)) f(α)=∫((απ+2ln 2)/(4(α^2 +1)))dα−∫((ln (α+1))/(α^2 +1))dα= =(π/8)ln (α^2 +1) +((ln 2)/2)tan^(−1) α −∫((ln (α+1))/(α^2 +1))dα We′re looking for f(1) ⇒ f(1)=∫_0 ^1 ((ln (x+1))/(x^2 +1))dx=(π/4)ln (2) −∫_0 ^1 ((ln (α+1))/(α^2 +1))dα 2∫_0 ^1 ((ln (x+1))/(x^2 +1))dx=(π/4)ln (2) ∫_0 ^1 ((ln (x+1))/(x^2 +1))dx=(π/8)ln (2)](https://www.tinkutara.com/question/Q210199.png)
$${f}\left(\alpha\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\left(\alpha{x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:\:\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\:'\left(\alpha\right)=\frac{{d}}{{d}\alpha}\left[\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\left(\alpha{x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\right]= \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\partial}{\partial\alpha}\left[\frac{\mathrm{ln}\:\left(\alpha{x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\right]{dx}= \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{x}}{\left(\alpha{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}= \\ $$$$=\frac{\mathrm{1}}{\alpha^{\mathrm{2}} +\mathrm{1}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{x}+\alpha}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}−\frac{\alpha}{\alpha^{\mathrm{2}} +\mathrm{1}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{dx}}{\alpha{x}+\mathrm{1}}= \\ $$$$=\left[\frac{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\alpha\mathrm{tan}^{−\mathrm{1}} \:{x}}{\alpha^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{ln}\:\left(\alpha{x}+\mathrm{1}\right)}{\alpha^{\mathrm{2}} +\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$$$=\frac{\alpha\pi+\mathrm{2ln}\:\mathrm{2}}{\mathrm{4}\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{ln}\:\left(\alpha+\mathrm{1}\right)}{\alpha^{\mathrm{2}} +\mathrm{1}} \\ $$$${f}\left(\alpha\right)=\int\frac{\alpha\pi+\mathrm{2ln}\:\mathrm{2}}{\mathrm{4}\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)}{d}\alpha−\int\frac{\mathrm{ln}\:\left(\alpha+\mathrm{1}\right)}{\alpha^{\mathrm{2}} +\mathrm{1}}{d}\alpha= \\ $$$$=\frac{\pi}{\mathrm{8}}\mathrm{ln}\:\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)\:+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\alpha\:−\int\frac{\mathrm{ln}\:\left(\alpha+\mathrm{1}\right)}{\alpha^{\mathrm{2}} +\mathrm{1}}{d}\alpha \\ $$$$\mathrm{We}'\mathrm{re}\:\mathrm{looking}\:\mathrm{for}\:{f}\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${f}\left(\mathrm{1}\right)=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\left({x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\frac{\pi}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{2}\right)\:−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left(\alpha+\mathrm{1}\right)}{\alpha^{\mathrm{2}} +\mathrm{1}}{d}\alpha \\ $$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\left({x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\frac{\pi}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{2}\right) \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{ln}\:\left({x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\frac{\pi}{\mathrm{8}}\mathrm{ln}\:\left(\mathrm{2}\right) \\ $$