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Question Number 210354 by klipto last updated on 08/Aug/24

∫_0 ^𝛂 (x/((1+x^2 )(1+𝛂x)))dx

$$\int_{\mathrm{0}} ^{\boldsymbol{\alpha}} \frac{\boldsymbol{\mathrm{x}}}{\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{1}+\boldsymbol{\alpha\mathrm{x}}\right)}\boldsymbol{\mathrm{dx}} \\ $$

Answered by klipto last updated on 08/Aug/24

$$ \\ $$

Answered by Frix last updated on 08/Aug/24

∫(x/((x^2 +1)(αx+1)))dx= =(1/(α^2 +1))∫(x/(x^2 +1))dx+(α/(α^2 +1))∫(dx/(x^2 +1))−(α/(α^2 +1))∫(dx/(αx+1))= =((ln (x^2 +1))/(2(α^2 +1)))+((αtan^(−1) x)/(α^2 +1))−((ln ∣αx+1∣)/(α^2 +1))+C ∫_0 ^α (x/((x^2 +1)(αx+1)))dx= =((αtan^(−1) α)/(α^2 +1))−((ln (α^2 +1))/(2(α^2 +1)))

$$\int\frac{{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\alpha{x}+\mathrm{1}\right)}{dx}= \\ $$$$=\frac{\mathrm{1}}{\alpha^{\mathrm{2}} +\mathrm{1}}\int\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\frac{\alpha}{\alpha^{\mathrm{2}} +\mathrm{1}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\alpha}{\alpha^{\mathrm{2}} +\mathrm{1}}\int\frac{{dx}}{\alpha{x}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\alpha\mathrm{tan}^{−\mathrm{1}} \:{x}}{\alpha^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{ln}\:\mid\alpha{x}+\mathrm{1}\mid}{\alpha^{\mathrm{2}} +\mathrm{1}}+{C} \\ $$$$\underset{\mathrm{0}} {\overset{\alpha} {\int}}\frac{{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left(\alpha{x}+\mathrm{1}\right)}{dx}= \\ $$$$=\frac{\alpha\mathrm{tan}^{−\mathrm{1}} \:\alpha}{\alpha^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{ln}\:\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)} \\ $$

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