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Question Number 51825 by Abdo msup. last updated on 30/Dec/18

$$findf\left(\alpha \right)={\int }_0^{1}ln(1+e^{-\alpha }x)dxwith\alpha ⩾0$$

Commented by maxmathsup by imad last updated on 31/Dec/18

$$letput{e}^{-\alpha }=\lambda \Rightarrow f\left(\alpha \right)={\int }_0^{1}ln(1+\lambda x)dxandbyparts{u}^{{}^{\prime }}=1,v=ln(1+\lambda x)$$$$wegetf\left(\alpha \right)={\left[xln(1+\lambda x)\right]}_0^1-{\int }_0^{1}x\frac{\lambda }{1+\lambda x}dx$$$$=ln(1+\lambda )-{\int }_0^1\frac{\lambda x+1-1}{1+\lambda x}dx=ln(1+\lambda )-1+{\int }_0^1\frac{dx}{1+\lambda x}$$$$=ln(1+\lambda )-1+\frac{1}{\lambda }ln\mid 1+\lambda x\mid +c\Rightarrow$$$$f\left(\alpha \right)=ln(1+e^{-\alpha })-1+e^{-\alpha }ln\mid 1+e^{-\alpha }x\mid +c.$$

Commented by maxmathsup by imad last updated on 31/Dec/18

$$f\left(\alpha \right)=ln(1+e^{-\alpha })-1+e^{\alpha }ln\mid 1+e^{-\alpha }x\mid +c.$$

Commented by Abdo msup. last updated on 31/Dec/18

$$f\left(\alpha \right)=ln(1+\lambda )-1+{[\frac{1}{\lambda }ln\mid 1+\lambda x\mid ]}_0^1+c$$$$=ln(1+\lambda )-1+\frac{1}{\lambda }ln(1+\lambda )+c$$$$\Rightarrow f\left(\alpha \right)=ln(1+e^{-\alpha })-1+e^{\alpha }ln(1+e^{-\alpha })+c$$$$=(1+e^{\alpha })ln(1+e^{-\alpha })-1+c.$$

Answered by Smail last updated on 30/Dec/18

$$lett=1+e^{-\alpha }x\Rightarrow dx=e^{\alpha }dt$$$$f\left(\alpha \right)=e^{\alpha }{\int }_1^{1+e^{-\alpha }}ln\left(t\right)dt$$$$byparts$$$$u=ln\left(t\right)\Rightarrow u^{\prime }=\frac{1}{t}$$$$v^{\prime }=1\Rightarrow v=t$$$$f\left(\alpha \right)=e^{\alpha }{\left[tlnt\right]}_1^{1+e^{-\alpha }}-e^{\alpha }{\int }_1^{1+e^{-\alpha }}dt$$$$=e^{\alpha }((1+e^{-\alpha })ln(1+e^{-\alpha })-e^{-\alpha })$$$$f\left(\alpha \right)=(e^{\alpha }+1)(ln(e^{\alpha }+1)-\alpha )-1$$

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