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Question Number 142365 by mathmax by abdo last updated on 30/May/21

$$\mathrm{calculate}\int \sqrt{1+{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{2\mathrm{x}}}\mathrm{dx}$$

Answered by MJS_new last updated on 05/Jun/21

$$\int \sqrt{{\mathrm{e}}^{2x}+{\mathrm{e}}^x+1}dx=$$$$[t={\mathrm{e}}^x+\frac{1}{2}\to dx=\frac{dt}{{\mathrm{e}}^x}]$$$$=\int \frac{\sqrt{4t^2+3}}{2t-1}dt=$$$$[u=\frac{2t+\sqrt{4t^2+3}}{\sqrt{3}}\to dt=\frac{\sqrt{4t^2+3}}{2u}du]$$$$=\frac{3\sqrt{3}}{4}\int \frac{{(u^2+1)}^2}{u^2(u-\sqrt{3})(3u+\sqrt{3})}du=$$$$=\int (\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4u^2}+\frac{1}{2u}+\frac{1}{u-\sqrt{3}}-\frac{3}{3u+\sqrt{3}})du=$$$$=\frac{\sqrt{3}u}{4}+\frac{\sqrt{3}}{4u}+\frac{\ln u}{2}+\ln (u-\sqrt{3})-\ln (3u+\sqrt{3})=$$$$=\frac{\sqrt{3}(u^2+1)}{4u}+\frac{1}{2}\ln \frac{u{(u-\sqrt{3})}^2}{{(3u+\sqrt{3})}^2}=$$$$...$$$$=\sqrt{{\mathrm{e}}^{2x}+{\mathrm{e}}^x+1}-x+\frac{1}{2}\ln ({2\mathrm{e}}^{3x}-{3\mathrm{e}}^{2x}-8+2({\mathrm{e}}^{2x}-{2\mathrm{e}}^x+4)\sqrt{{\mathrm{e}}^{2x}+{\mathrm{e}}^x+1})+C$$

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