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Question Number 96346 by Ar Brandon last updated on 31/May/20

$$\int \frac{1}{\mathfrak{ln}\left(\mathfrak{x}\right)}\mathbf{d}\mathfrak{x}$$

Commented by Tony Lin last updated on 01/Jun/20

$$letlnx=t,dx=xdt=e^{t}dt$$$$\int \frac{e^t}{t}dt$$$$=Ei\left(t\right)+c$$$$=Ei\left(lnx\right)+c$$

Commented by Sourav mridha last updated on 01/Jun/20

$$\mathit{w}\mathit{r}\mathit{o}\mathit{n}\mathit{g}\mathit{d}\mathit{e}\mathit{f}\mathit{i}\mathit{n}\mathit{e}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{o}\mathit{f}\mathit{e}\mathit{x}\mathit{p}\mathit{o}\mathit{n}\mathit{e}\mathit{n}\mathit{t}\mathit{i}\mathit{a}\mathit{l}$$$$\mathit{i}\mathit{n}\mathit{t}\mathit{e}\mathit{g}\mathit{r}\mathit{a}\mathit{l}...$$$$\mathit{t}\mathit{h}\mathit{i}\mathit{s}\mathit{i}\mathit{s}\mathit{l}\mathit{i}\mathit{k}\mathit{e}\mathit{t}\mathit{h}\mathit{a}\mathit{t}$$$$-\mathit{E}\mathrm{i}(-\mathrm{x})\equiv {\int }_{\mathrm{x}}^{\mathrm{\infty }}\frac{{\mathit{e}}^{-\mathit{t}}}{\mathit{t}}\mathit{d}\mathit{t}\equiv {\mathit{E}}_1\left(\mathit{x}\right).$$

Commented by Tony Lin last updated on 02/Jun/20

$$Ei\left(x\right)={\int }_{-\mathrm{\infty }}^x\frac{e^t}{t}dt$$$$\to \int \frac{e^x}{x}dx=Ei\left(x\right)+c$$

Commented by Tony Lin last updated on 02/Jun/20

$$\int \frac{1}{lnx}dx$$$$=li\left(x\right)+c$$

Answered by niroj last updated on 01/Jun/20

$$\int \frac{1}{\mathfrak{ln}\left(\mathfrak{x}\right)}\mathbf{d}\mathfrak{x}=\int \frac{1}{\ln \left(\mathrm{x}\right)}\mathrm{dx}$$$$\mathrm{Put},\mathrm{In}\mathrm{x}=\mathrm{t}$$$$\mathrm{x}={\mathrm{e}}^{\mathrm{t}}$$$$\mathrm{dx}={\mathrm{e}}^{\mathrm{t}}\mathrm{dt}$$$$\int \frac{1}{\mathrm{t}}.{\mathrm{e}}^{\mathrm{t}}\mathrm{dt}$$$$=\int \frac{1}{\mathrm{t}}(1+\frac{\mathrm{t}}{1!}+\frac{{\mathrm{t}}^2}{2!}+\frac{{\mathrm{t}}^3}{3\cdot }+...)\mathrm{dt}$$$$=\int (\frac{1}{\mathrm{t}}+1+\frac{\mathrm{t}}{2!}+\frac{{\mathrm{t}}^2}{3!}+....)\mathrm{dt}$$$$=\log \mathrm{t}+\mathrm{t}+\frac{1}{2}.\frac{{\mathrm{t}}^2}{2}+\frac{1}{6}.\frac{{\mathrm{t}}^3}{3}+...\mathrm{C}$$$$=\log \mathrm{t}+\mathrm{t}+\frac{1}{4}{\mathrm{t}}^2+\frac{1}{18}{\mathrm{t}}^3+....\mathrm{C}$$$$\int \frac{1}{\mathrm{In}\left(\mathrm{x}\right)}\mathrm{dx}=\log \left(\mathrm{In}\mathrm{x}\right)+\ln \mathrm{x}+\frac{1}{4}{\left(\ln \mathrm{x}\right)}^2+\frac{1}{18}{\left(\ln \mathrm{x}\right)}^3+...\mathrm{C}$$$$\int \frac{1}{\ln \left(\mathfrak{x}\right)}\mathrm{d}\mathfrak{x}=\log \left(\ln \mathfrak{x}\right)+\mathrm{In}\mathfrak{x}+\frac{1}{4}{\left(\ln \mathfrak{x}\right)}^2+\frac{1}{18}{\left(\mathrm{In}\mathfrak{x}\right)}^3+...\mathrm{C}//.$$$$$$$$$$

Commented by Ar Brandon last updated on 01/Jun/20

Thank you Mr niroj, ��

Commented by Mr.D.N. last updated on 01/Jun/20

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