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Question Number 112313 by mathdave last updated on 07/Sep/20

$$solve$$$${\int }_0^1\frac{\ln x}{\sqrt{1+x^2}}dx$$

Commented by mohammad17 last updated on 07/Sep/20

$$set:x=tan\left(r\right)\Rightarrow dx={sec}^2\left(r\right)dr$$$$x=0\Rightarrow r=0,x=1\Rightarrow r=\frac{\pi }{4}$$$$$$$${\int }_0^{\frac{\pi }{4}}{sec}^3\left(r\right)dr$$$$$$$$byusingtherule$$$$$$$$\int {sec}^n\left(x\right)dx=\frac{{sec}^{n-1}\left(x\right)}{(n-1)csc\left(x\right)}+\frac{(n-2)}{(n-1)}\int {sec}^{n-2}\left(x\right)dx$$$$$$$$\text{Missing left or extra right}$$$$$$$$Note:\int sec\left(r\right)dr=ln\mid sec\left(r\right)+tan\left(r\right)\mid +C$$$$$$$$\therefore {\int }_0^1\sqrt{1+x^2}dx=\frac{1}{2}\left(\sqrt{2}\right)+\frac{1}{2}ln\mid \sqrt{2}+1\mid =\frac{\sqrt{2}+ln\mid 1+\sqrt{2}\mid }{2}$$

Commented by mathdave last updated on 07/Sep/20

$$amsososorry{sir}^{\prime }stochangethephaseofthe$$$$question.theoneitypebeforeieluded$$$$toinclude\left(\ln x\right)ontopof\sqrt{1+x^2}.allworks$$$$uguysdidinitiallywasallcorrecti$$$$cantdenythatfact.uguysregreat$$$$mathematicians$$

Answered by Dwaipayan Shikari last updated on 07/Sep/20

$${[\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}log(x+\sqrt{x^2+1})]}_0^1=\frac{1}{\sqrt{2}}+\frac{1}{2}log(1+\sqrt{2})$$

Answered by MJS_new last updated on 07/Sep/20

$$\underset{0}{\stackrel{1}{\int }}\sqrt{x^2+1}dx=$$$$[t=x+\sqrt{x^2+1}\to dx=\frac{\sqrt{x^2+1}}{x+\sqrt{x^2+1}}dt]$$$$=\underset{1}{\stackrel{1+\sqrt{2}}{\int }}(\frac{t}{4}+\frac{1}{2t}+\frac{1}{4t^3})dt=$$$$={[\frac{t^2}{8}+\frac{\ln t}{2}-\frac{1}{8t^2}]}_1^{1+\sqrt{2}}=\frac{\sqrt{2}+\ln (1+\sqrt{2})}{2}$$

Answered by mathdave last updated on 07/Sep/20

$$solutionto{\int }_0^1\frac{\ln x}{\sqrt{1+x^2}}dx$$$$letx=\sinh \theta ,dx=\cosh \theta d\theta and$$$${\sinh }^{-1}\left(x\right)=\ln (x+\sqrt{1+x^2})$$$$I={\int }_{{\sinh }^{-1}\left(0\right)}^{{\sinh }^{-1}\left(1\right)}\frac{\ln \left(\sinh \theta \right)}{\cosh \theta }d\theta ={\int }_0^{{\sinh }^{-1}\left(1\right)}\ln \left(\sinh \theta \right)d\theta$$$$but\sinh \theta =\frac{e^{\theta }-e^{-\theta }}{2}$$$$I={\int }_0^{\ln (1+\sqrt{2})}\ln (e^{\theta }-e^{-\theta })d\theta -{\int }_0^{\ln (1+\sqrt{2})}\ln \left(2\right)d\theta$$$$I={\int }_0^{\ln (1+\sqrt{2})}\ln \left(e^{\theta }\right)d\theta +{\int }_0^{\ln (1+\sqrt{2})}\ln (1-e^{-\theta })d\theta -{\int }_0^{\ln (1+\sqrt{2})}\ln \left(2\right)d\theta$$$$I=\frac{{\ln }^2(1+\sqrt{2})}{2}-{\int }_0^{\ln (1+\sqrt{2})}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{e^{-2(k+1)\theta }}{k+1}d\theta -\ln \left(2\right)\ln (1+\sqrt{2})$$$$I=\frac{{\ln }^2(1+\sqrt{2})}{2}+\frac{1}{2}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{k+1}\left[\frac{e^{-2(k+1)\ln (1+\sqrt{2})}-1}{k+1}\right]-\ln \left(2\right)\ln (1+\sqrt{2})$$$$I=\frac{{\ln }^2(1+\sqrt{2})}{2}+\frac{1}{2}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(1+\sqrt{2})}^{-2{(k+1)}_{}}}{{(k+1)}^2}-\frac{1}{2}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(k+1)}^2}-\ln \left(2\right)\ln (1+\sqrt{2})$$$$I=\frac{{\ln }^2(1+\sqrt{2})}{2}+\frac{1}{2}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{\frac{{(1+\sqrt{2})}^{2k}}{k^2}}-\frac{1}{2}\zeta 2)-\ln \left(2\right)\ln (1+\sqrt{2})$$$$but{Li}_2\left(x\right)=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{x^k}{k^2}$$$$I=\frac{{\ln }^2(1+\sqrt{2})}{2}+\frac{1}{2}{Li}_2\left(\frac{1}{{(1+\sqrt{2})}^2}\right)-\frac{1}{2}\left(\frac{{\pi }^2}{6}\right)-\ln \left(2\right)\ln (1+\sqrt{2})$$$$I=\frac{{\ln }^2(1+\sqrt{2})}{2}+\frac{1}{2}{Li}_2\left(\frac{1}{3+2\sqrt{2}}\right)-\frac{{\pi }^2}{12}-\ln \left(2\right)\ln (1+\sqrt{2})$$$$I=\frac{{\ln }^2(1+\sqrt{2})}{2}+\frac{1}{2}{Li}_2(3-2\sqrt{2})-\frac{{\pi }^2}{12}-\ln \left(2\right)\ln (1+\sqrt{2})$$$$I=\frac{{\ln }^2(1+\sqrt{2})}{2}-\frac{{\pi }^2}{12}-\ln \left(2\right)\ln (1+\sqrt{2})+\frac{1}{2}{Li}_2(3-2\sqrt{2})$$$$\because {\int }_0^1\frac{\ln x}{\sqrt{1+x^2}}dx=\frac{{\ln }^2(1+\sqrt{2})}{2}-\frac{{\pi }^2}{12}-\ln \left(2\right)\ln (1+\sqrt{2})+\frac{1}{2}{Li}_2(3-2\sqrt{2})$$$$bymathdave\left(07/09/2020\right)$$

Commented by Tinku Tara last updated on 08/Sep/20

$$\mathrm{mathdave},\mathrm{please}\mathrm{stop}\mathrm{using}\mathrm{abusive}$$$$\mathrm{language}\mathrm{on}\mathrm{this}\mathrm{forum}\mathrm{immediately}.$$

Commented by Her_Majesty last updated on 07/Sep/20

$$\int \sqrt{1+x^2}dxwithx=sinhtleadsto$$$$\int {cosh}^{2}tdt=\frac{1}{2}(t+coshtsinht)=$$$$=\frac{1}{2}({sinh}^{-1}x+x\sqrt{x^2+1})+C$$$$noneedforsuchalongtourthroughthe$$$$lowlandsofmath$$

Commented by MJS_new last updated on 07/Sep/20

$$\mathrm{you}\mathrm{cannot}\mathrm{hold}\mathrm{it}\mathrm{back}\mathrm{it}\mathrm{seems}.\mathrm{sad}.$$

Commented by MJS_new last updated on 07/Sep/20

$$\mathrm{and}\mathrm{you}\mathrm{too}\mathrm{cannot}\mathrm{hold}\mathrm{it}\mathrm{back}.$$$$\mathrm{the}\mathrm{two}\mathrm{of}\mathrm{you}\mathrm{are}\mathrm{perfect}\mathrm{mirrors}\mathrm{for}\mathrm{one}$$$$\mathrm{another}.$$

Commented by mathdave last updated on 07/Sep/20

$$ntthaticantholdbackbutsheisdoing$$$$hasifshemonopolyofknowledge$$

Commented by MJS_new last updated on 07/Sep/20

$$\mathrm{Mr}\mathrm{Mathdave},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{exactly}\mathrm{the}\mathrm{same}\mathrm{with}$$$$\mathrm{you}.\mathrm{scroll}\mathrm{down}\mathrm{and}\mathrm{read}\mathrm{your}\mathrm{own}\mathrm{comments}!$$

Commented by mathdave last updated on 07/Sep/20

$$hasuhavetherighttostayindforum$$$$datishowihavethefullrighttostay$$$$onthisforumbcosnobodyis$$$$subscribingforanybody$$

Commented by mathdave last updated on 07/Sep/20

$$sorrymrmjsdquestionwasatypo,i$$$$alreadycorrectedit$$

Commented by mathdave last updated on 07/Sep/20

$$ifudontlikeuredonethatwillleave$$$$formeindsgroup$$

Commented by ajfour last updated on 07/Sep/20

$$stayintheforum,withsomedecent$$$$conversationandwords,ihavebeen$$$$ontheforumforsomeyesrs;never$$$$witnessedanyoneusingabusive$$$$words..kindlyyoutoo,abidebythe$$$$samedecorum..$$

Commented by MJS_new last updated on 07/Sep/20

$$\mathrm{The}\mathrm{question}{\mathrm{wasn}}^{\prime }\mathrm{t}\mathrm{a}\mathrm{typo},\mathrm{you}\mathrm{have}\mathrm{done}$$$$\mathrm{this}\mathrm{before}.\mathrm{You}\mathrm{want}\mathrm{to}\mathrm{make}\mathrm{others}\mathrm{look}$$$$\mathrm{like}\mathrm{fools}\mathrm{but}\mathrm{most}\mathrm{of}\mathrm{us}\mathrm{here}\mathrm{have}\mathrm{become}$$$$\mathrm{friends}\mathrm{over}\mathrm{the}\mathrm{years}...\mathrm{no}\mathrm{more}\mathrm{to}\mathrm{tell}\mathrm{you},$$$${\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{ignore}\mathrm{you}\mathrm{from}\mathrm{now}\mathrm{on}.$$

Commented by mathmax by abdo last updated on 07/Sep/20

$$\mathrm{how}\mathrm{are}\mathrm{sir}\mathrm{mjs}?$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

Answered by mathmax by abdo last updated on 07/Sep/20

$$\mathrm{I}={\int }_0^1\sqrt{1+{\mathrm{x}}^2}\mathrm{dx}\mathrm{by}\mathrm{parts}\mathrm{I}={\left[\mathrm{x}\sqrt{1+{\mathrm{x}}^2}\right]}_0^1-{\int }_0^1\mathrm{x}\frac{2\mathrm{x}}{2\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx}$$$$=\sqrt{2}-{\int }_0^1\frac{{\mathrm{x}}^2}{\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx}=\sqrt{2}-{\int }_0^1\frac{1+{\mathrm{x}}^2-1}{\sqrt{1+{\mathrm{x}}^2}}\mathrm{dx}$$$$=\sqrt{2}-{\int }_0^1\sqrt{1+{\mathrm{x}}^2}+{\int }_0^1\frac{\mathrm{dx}}{\sqrt{1+{\mathrm{x}}^2}}\Rightarrow 2\mathrm{I}=\sqrt{2}+[\ln {(\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}]}_0^1$$$$=\sqrt{2}+\ln (1+\sqrt{2})\Rightarrow \mathrm{I}=\frac{\sqrt{2}}{2}+\frac{1}{2}\ln (1+\sqrt{2})$$

Answered by 1549442205PVT last updated on 10/Sep/20

$$\mathrm{Put}\frac{1}{{\mathrm{x}}^2}+1={\mathrm{t}}^2\Rightarrow -\frac{2\mathrm{dx}}{{\mathrm{x}}^3}=2\mathrm{tdt}$$$$\Rightarrow \mathrm{dx}=-{\mathrm{x}}^3\mathrm{tdt},\sqrt{1+{\mathrm{x}}^2}=\mathrm{xt}.\mathrm{Hence},$$$$\mathrm{F}={\int }_0^1\sqrt{1+x^2}dx={\int }_{\sqrt{2}}^{\mathrm{\infty }}{\mathrm{x}}^4{\mathrm{t}}^2\mathrm{dt}={\int }_{\sqrt{2}}^{\mathrm{\infty }}\frac{{\mathrm{t}}^2\mathrm{dt}}{{({\mathrm{t}}^2-1)}^2}$$$$={\int }_{\sqrt{2}}^{\mathrm{\infty }}\frac{({\mathrm{t}}^2-1)+1}{{({\mathrm{t}}^2-1)}^2}\mathrm{dt}={\int }_{\sqrt{2}}^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2-1}+{\int }_{\sqrt{2}}^{\mathrm{\infty }}\frac{\mathrm{dt}}{{({\mathrm{t}}^2-1)}^2}$$$$={\int }_{\sqrt{2}}^{\mathrm{\infty }}(\mathrm{A}+\mathrm{B})$$$$\mathrm{A}=\frac{1}{{\mathrm{t}}^2-1}=\frac{1}{2}(\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}+1}).\mathrm{Put}\frac{1}{\mathrm{t}-1}=\mathrm{a},\frac{1}{\mathrm{t}+1}=\mathrm{b}$$$$\Rightarrow \mathrm{A}=\mathrm{ab}=\frac{1}{2}(\mathrm{a}-\mathrm{b})$$$$\mathrm{we}\mathrm{get}\mathrm{B}={\mathrm{A}}^2={\left[\frac{1}{2}(\mathrm{a}-\mathrm{b})\right]}^2=\frac{1}{4}({\mathrm{a}}^2+{\mathrm{b}}^2)$$$$-\frac{1}{2}\mathrm{ab}=\frac{1}{4}({\mathrm{a}}^2+{\mathrm{b}}^2)-\frac{1}{4}(\mathrm{a}-\mathrm{b}).\mathrm{Hence}$$$$\Rightarrow \mathrm{A}+\mathrm{B}=\frac{1}{4}({\mathrm{a}}^2+{\mathrm{b}}^2)+\frac{1}{4}(\mathrm{a}-\mathrm{b})=\frac{1}{4}[\frac{1}{{(\mathrm{t}-1)}^2}+\frac{1}{{(\mathrm{t}+1)}^2}]+\frac{1}{4}(\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}+1})$$$$\mathrm{F}={\int }_{\sqrt{2}}^{\mathrm{\infty }}(\mathrm{A}+\mathrm{B})=\frac{1}{4}{\int }_{\sqrt{2}}^{\mathrm{\infty }}\frac{\mathrm{dt}}{{(\mathrm{t}-1)}^2}+\frac{1}{4}{\int }_{\sqrt{2}}^{\mathrm{\infty }}\frac{\mathrm{dt}}{{(\mathrm{t}+1)}^2}$$$$+\frac{1}{4}{\int }_{\sqrt{2}}^{\mathrm{\infty }}(\frac{1}{\mathrm{t}-1}-\frac{1}{\mathrm{t}+1})\mathrm{dt}$$$$={[-\frac{1}{4(\mathrm{t}-1)}-\frac{1}{4(\mathrm{t}+1)}]}_{\sqrt{2}}^{\mathrm{\infty }}+\frac{1}{4}\ln \mid \frac{\mathrm{t}-1}{\mathrm{t}+1}{\mid }_{\sqrt{2}}^{\mathrm{\infty }}$$$$=-\frac{1}{4(\sqrt{2}-1)}-\frac{1}{4(\sqrt{2}+1)}-\frac{1}{4}\ln \frac{\sqrt{2}-1}{\sqrt{2}+1}$$$$=\frac{1}{4}(\sqrt{2}+1+\sqrt{2}-1)-\frac{1}{4}\ln \left(\frac{1}{{(\sqrt{2}+1)}^2}\right)$$$$=\frac{\sqrt{2}}{2}+\frac{1}{2}\ln (\sqrt{2}+1)$$

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