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Question Number 110619 by mathdave last updated on 29/Aug/20

$$someonepostedthisproblemandmy$$$$solutionfollowed$$$$solve$$$${\int }_0^1\frac{\ln (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx$$$$solution$$$$y=\ln (x+\sqrt{1+x^2})anddy=\frac{1}{\sqrt{1+x^2}}dx$$$$atx=0,y=0andatx=1,y=\ln (1+\sqrt{2})$$$$hence$$$$I={\int }_0^{\ln (1+\sqrt{2})}\frac{y}{\sqrt{1+x^2}}\times \sqrt{1+x^2}dy={\int }_0^{\ln (1+\sqrt{2})}ydy$$$$I={\left[\frac{y^2}{2}\right]}_0^{\ln (1+\sqrt{2})}=\frac{1}{2}{\ln }^2(1+\sqrt{2})$$$$\because {\int }_0^1\frac{\ln (x+\sqrt{1+x^2})}{\sqrt{1+x^2}}dx=\frac{1}{2}{\ln }^2(1+\sqrt{2})$$

Commented by mathdave last updated on 29/Aug/20

$$whydidyoumarkitred.whatisthat$$$$supposetomean$$

Commented by Her_Majesty last updated on 30/Aug/20

$$I{didn}^{\prime }tmarkitredbutIcantellyouthe$$$$reasonsomeonedid:$$$$wehave3optionstopost$$$$\left(1\right)questions$$$$\left(2\right)answers$$$$\left(3\right)comments$$$$$$$$\left(1\right)istopostquestions$$$$\left(2\right)istopostanswers$$$$\left(3\right)istopostcomments$$$$$$$$donotpostquestionsin\left(2\right)or\left(3\right)$$$$donotpostanswersin\left(1\right)or\left(3\right)$$$$donotpostcommentsin\left(1\right)or\left(2\right)$$$$$$$${it}^{\prime }seasy,{isn}^{\prime }tit?$$

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