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Question Number 39135 by maxmathsup by imad last updated on 02/Jul/18

$$calculateA\left(\lambda \right)={\int }_0^{\lambda }\frac{ln(x+\sqrt{1+x^2)}}{\sqrt{1+x^2}}dx$$$$2)calculate{\int }_0^1\frac{ln(x+\sqrt{1+x^2)}}{\sqrt{1+x^2}}dx$$

Commented by math khazana by abdo last updated on 03/Jul/18

$$letintegratebyparts{u}^{{}^{\prime }}=\frac{1}{\sqrt{1+x^2}}andv=ln(x+\sqrt{1+x^2)}$$$$A\left(\lambda \right)={\left[{ln}^2(x+\sqrt{1+x^2})\right]}_0^{\lambda }-{\int }_0^{\lambda }\frac{ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$$$$\Rightarrow 2A\left(\lambda \right)={ln}^2(\lambda +\sqrt{1+{\lambda }^2})\Rightarrow$$$$A\left(\lambda \right)=\frac{1}{2}{ln}^2(\lambda +\sqrt{1+{\lambda }^2})$$$$2){\int }_0^1\frac{ln(1+\sqrt{1+x^2}}{\sqrt{1+x^2}})dx=A\left(1\right)=\frac{1}{2}{ln}^2(2+\sqrt{2}).$$

Commented by math khazana by abdo last updated on 03/Jul/18

$$A\left(1\right)=\frac{1}{2}{ln}^2(1+\sqrt{2})$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jul/18

$$1)t=ln(x+\sqrt{1+x^2})$$$$\frac{dt}{dx}=\frac{1+\frac{2x}{2\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=\frac{1}{\sqrt{1+x^2}}$$$${\int }_0^{ln(\lambda +\sqrt{1+{\lambda }^2})}tdt$$$$\mid \frac{t^2}{2}{\mid }_0^{ln(\lambda +\sqrt{1+{\lambda }^2)}}$$$$\frac{1}{2}\{ln{(\lambda +\sqrt{1+{\lambda }^2)}\}}^2$$$$2)\frac{1}{2}{\{ln\mid 1+\sqrt{2}\mid \}}^2$$

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