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Question Number 49003 by mondodotto@gmail.com last updated on 01/Dec/18

$$\int \frac{\mathbf{log}\mathit{x}}{\sqrt{1-{\mathit{x}}^{\mathit{x}}}}\mathbf{d}\mathit{x}$$$$\mathbf{please}\mathbf{help}\mathbf{this}$$

Commented by maxmathsup by imad last updated on 01/Dec/18

$$I=\int \frac{ln\left(x\right)}{\sqrt{1-e^{xln\left(x\right)}}}dxchangementxln\left(x\right)=tgive(ln\left(x\right)+1)dx=dt$$$$I=\int \frac{ln\left(x\right)+1-1}{\sqrt{1-x^x}}dx=\int \frac{ln\left(x\right)+1}{\sqrt{1-e^{xlnx}}}dx-\int \frac{dx}{\sqrt{1-x^x}}$$$$=\int \frac{dt}{\sqrt{1-t}}-\int \frac{dx}{\sqrt{1-x^x}}=-2\sqrt{1-t}-\int \frac{dx}{\sqrt{1-x^x}}=-2\sqrt{1-x^x}-\int \frac{dx}{\sqrt{1-x^x}}$$$$letfind\int \frac{dx}{\sqrt{1-x^x}}atformofseriedueto0<x^x<1$$$$wehave{(1+u)}^{\alpha }=1+\frac{\alpha u}{1!}+\frac{\alpha (\alpha -1)}{2!}{u}^2+....+\frac{\alpha (\alpha -1)...(\alpha -n+1)}{n!}{u}^n+...\Rightarrow$$$${(1-u)}^{\alpha }=1-\alpha u+\frac{\alpha (\alpha -1)}{2!}{u}^2+\frac{\alpha (\alpha -1)...(\alpha -n+1)}{n!}{(-1)}^n{u}^n+...\Rightarrow$$$${(1-x^x)}^{-\frac{1}{2}}=1+\frac{1}{2}{x}^x+\frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}{x}^{2x}+...\frac{(-\frac{1}{2})(-\frac{3}{2})...(-\frac{1}{2}-n+1)}{n!}{(-1)}^n{x}^{nx}+...$$$$\Rightarrow \int \frac{dx}{\sqrt{1-x^x}}=x+\frac{1}{2}\int x^{x}dx+\int \frac{3}{42!}{x}^{2x}dx+...$$$$+\frac{(-\frac{1}{2})(-\frac{3}{2})....(-\frac{1}{2}-n+1)}{n!}{(-1)}^n\int x^{nx}dx+....$$

Commented by MJS last updated on 01/Dec/18

$$\mathrm{where}\mathrm{did}\mathrm{you}\mathrm{find}\mathrm{this}?$$

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