Question Number 183687 by Mastermind last updated on 28/Dec/22
$$\int\frac{\mathrm{1}}{\mathrm{lnx}}\mathrm{dx} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}\:\mathrm{out} \\ $$
Answered by a.lgnaoui last updated on 28/Dec/22
![(1/(ln(x)))=((xln(x))/(x[ln(x)]^2 ))=(((lnx)/x))((x/([ln(x)]^2 ))) =d(lnx)lnx(x/([ln(x)]^2 ))=((xd[(lnx)^2 ])/(2[ln(x)]^2 )) posons u=[ln(x)]^2 v=(x/([ln(x)]^2 )) V′=(([ln(x)]^2 −2lnx)/([ln(x)]^2 )) I=(1/2)[(x)−∫[ln(x)^2 ]−2ln(x)]dx] =(1/2)[x−∫[(lnx−1)^2 −1]dx] = x −(1/2) ∫(lnx−1)^2 dx posons P=(lnx−1)^2 dP=(2/x)(lnx−1) dQ=1 Q=x ∫(lnx−1)^2 dx=x(lnx−1)^2 −2∫(lnx−1)dx =x(lnx−1)^2 +4x−2xlnx+c soit: I=x−(1/2)[x(lnx−1)^2 −2xlnx+4x+C Conclusion: ∫(dx/(lnx)) = x[lnx−(1/2)(lnx−1)^2 −1]+C j](Q183697.png)
$$\:\:\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{x}\right)}=\frac{\mathrm{xln}\left(\mathrm{x}\right)}{\mathrm{x}\left[\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{2}} }=\left(\frac{\mathrm{lnx}}{\mathrm{x}}\right)\left(\frac{\mathrm{x}}{\left[\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{2}} }\right) \\ $$$$=\mathrm{d}\left(\mathrm{lnx}\right)\mathrm{lnx}\frac{\mathrm{x}}{\left[\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{2}} }=\frac{\mathrm{xd}\left[\left(\mathrm{lnx}\right)^{\mathrm{2}} \right]}{\mathrm{2}\left[\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{2}} } \\ $$$${posons}\:\:\mathrm{u}=\left[\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{2}} \:\:\:\:\:\mathrm{v}=\frac{\mathrm{x}}{\left[\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{V}'=\frac{\left[\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{2}} −\mathrm{2lnx}}{\left[\mathrm{ln}\left(\mathrm{x}\right)\right]^{\mathrm{2}} } \\ $$$$\left.\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{x}\right)−\int\left[\mathrm{ln}\left(\mathrm{x}\right)^{\mathrm{2}} \right]−\mathrm{2ln}\left(\mathrm{x}\right)\right]\mathrm{dx}\right]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{x}−\int\left[\left(\mathrm{lnx}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right]\mathrm{dx}\right] \\ $$$$\:\:\:=\:\:\mathrm{x}\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int\left(\mathrm{lnx}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{posons} \\ $$$$\mathrm{P}=\left(\mathrm{lnx}−\mathrm{1}\right)^{\mathrm{2}} \:\:\:\:\:\:\mathrm{dP}=\frac{\mathrm{2}}{\mathrm{x}}\left(\mathrm{lnx}−\mathrm{1}\right) \\ $$$$\mathrm{dQ}=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}=\mathrm{x} \\ $$$$\int\left(\mathrm{lnx}−\mathrm{1}\right)^{\mathrm{2}} \mathrm{dx}=\mathrm{x}\left(\mathrm{lnx}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\int\left(\mathrm{lnx}−\mathrm{1}\right)\mathrm{dx} \\ $$$$=\mathrm{x}\left(\mathrm{lnx}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4x}−\mathrm{2xlnx}+\mathrm{c} \\ $$$$\mathrm{soit}: \\ $$$${I}=\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{x}\left(\mathrm{lnx}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2xlnx}+\mathrm{4x}+\mathrm{C}\right. \\ $$$$\:\:{Conclusion}: \\ $$$$\:\:\:\int\frac{\mathrm{dx}}{\mathrm{lnx}}\:=\:\mathrm{x}\left[\mathrm{lnx}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{lnx}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right]+\mathrm{C} \\ $$$${j} \\ $$
Commented by Frix last updated on 28/Dec/22
$$\mathrm{Sorry}\:\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{nonsense}. \\ $$
Answered by Frix last updated on 28/Dec/22
![∫(dx/(ln x))=li x +C [by definition: li x = Integral Logarithm]](Q183698.png)
$$\int\frac{{dx}}{\mathrm{ln}\:{x}}=\mathrm{li}\:{x}\:+{C} \\ $$$$\left[\mathrm{by}\:\mathrm{definition}:\:\mathrm{li}\:{x}\:=\:\mathrm{Integral}\:\mathrm{Logarithm}\right] \\ $$
Answered by paul2222 last updated on 28/Dec/22
![let u=ln(x) x=e^u dx=e^u du ∫(e^u /u)du we recall e^u =𝚺_(n=0) ^∞ (u^n /(n!)) 𝚺_(n=0) ^∞ (1/(n!))∫u^n du=𝚺_(n=0) ^∞ (1/(n!))[(u^(n+1) /(n+1))]+C where u=ln(x) 𝚺_(n=0) ^∞ ((ln(x)^(n+1) )/(n!(n+1)))=ln(x)𝚺_(n=0) ^∞ (((ln(x))^n )/(n!(n+1)))+C](Q183705.png)
$$\boldsymbol{\mathrm{let}}\:\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{u}}} \\ $$$$\boldsymbol{\mathrm{dx}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{u}}} \boldsymbol{\mathrm{du}} \\ $$$$\int\frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{u}}} }{\boldsymbol{\mathrm{u}}}\boldsymbol{\mathrm{du}} \\ $$$$\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{recall}}\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{u}}} =\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\boldsymbol{\mathrm{u}}^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}!} \\ $$$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}!}\int\boldsymbol{\mathrm{u}}^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{du}}=\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}!}\left[\frac{\boldsymbol{\mathrm{u}}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\boldsymbol{\mathrm{n}}+\mathrm{1}}\right]+\boldsymbol{\mathrm{C}}\: \\ $$$$\boldsymbol{\mathrm{where}}\:\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\boldsymbol{\mathrm{n}}!\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)}=\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\right)^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}!\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)}+\boldsymbol{\mathrm{C}} \\ $$