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Question Number 201172 by Calculusboy last updated on 01/Dec/23

Answered by Sutrisno last updated on 01/Dec/23

$$=\int \frac{2e^{2x}-e^x}{\sqrt{3(e^{2x}-2e^x-\frac{1}{3})}}dx$$$$=\frac{1}{\sqrt{3}}\int \frac{2e^{2x}-e^x}{\sqrt{{(e^x-1)}^2-\frac{4}{3}}}dx$$$$misal:e^x-1=\frac{2}{\sqrt{3}}sec\theta \to tan\theta =\frac{\sqrt{3e^{2x}-6e^x-1}}{2}$$$$dx=\frac{2}{\sqrt{3}}.\frac{sec\theta .tan\theta }{e^x}d\theta =\frac{2}{\sqrt{3}}\left(\frac{sec\theta .tan\theta }{\frac{2}{\sqrt{3}}sec\theta +1}\right)d\theta$$$$=\frac{1}{\sqrt{3}}\int \frac{2{(\frac{2}{\sqrt{3}}sec\theta +1)}^2-(\frac{2}{\sqrt{3}}sec\theta +1)}{\sqrt{{\left(\frac{2}{\sqrt{3}}sec\theta \right)}^2-\frac{4}{3}}}.\frac{2}{\sqrt{3}}\left(\frac{sec\theta .tan\theta }{\frac{2}{\sqrt{3}}sec\theta +1}\right)d\theta$$$$=\frac{1}{\sqrt{3}}\int \frac{2(\frac{2}{\sqrt{3}}sec\theta +1)-1}{\frac{2}{\sqrt{3}}.tan\theta }.\frac{2}{\sqrt{3}}.sec\theta tan\theta d\theta$$$$=\frac{1}{\sqrt{3}}\int \frac{4}{\sqrt{3}}{sec}^2\theta +sec\theta d\theta$$$$=\frac{1}{\sqrt{3}}(\frac{4}{\sqrt{3}}tan\theta +ln\mid sec\theta +tan\theta \mid )$$$$=\frac{1}{\sqrt{3}}(\frac{4}{\sqrt{3}}\frac{\sqrt{3e^{2x}-6e^x-1}}{2}+ln\mid \frac{\sqrt{3}(e^x-1)}{2}+\frac{\sqrt{3e^{2x}-6e^x-1}}{2}\mid )+c$$$$$$$$$$

Commented by Calculusboy last updated on 04/Dec/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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