calculate-1-e-x-e-2x-dx-

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

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

Answered by MJS_new last updated on 05/Jun/21

\int \sqrt{e^{2 x} + e^{x} + 1} d x =

[t = e^{x} + \frac{1}{2} \to d x = \frac{d t}{e^{x}}]

= \int \frac{\sqrt{4 t^{2} + 3}}{2 t - 1} d t =

[u = \frac{2 t + \sqrt{4 t^{2} + 3}}{\sqrt{3}} \to d t = \frac{\sqrt{4 t^{2} + 3}}{2 u} d u]

= \frac{3 \sqrt{3}}{4} \int \frac{{(u^{2} + 1)}^{2}}{u^{2} (u - \sqrt{3}) (3 u + \sqrt{3})} d u =

= \int (\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4 u^{2}} + \frac{1}{2 u} + \frac{1}{u - \sqrt{3}} - \frac{3}{3 u + \sqrt{3}}) d u =

= \frac{\sqrt{3} u}{4} + \frac{\sqrt{3}}{4 u} + \frac{\ln u}{2} + \ln (u - \sqrt{3}) - \ln (3 u + \sqrt{3}) =

= \frac{\sqrt{3} (u^{2} + 1)}{4 u} + \frac{1}{2} \ln \frac{u {(u - \sqrt{3})}^{2}}{{(3 u + \sqrt{3})}^{2}} =

\dots

= \sqrt{e^{2 x} + e^{x} + 1} - x + \frac{1}{2} \ln ({2 e}^{3 x} - {3 e}^{2 x} - 8 + 2 (e^{2 x} - {2 e}^{x} + 4) \sqrt{e^{2 x} + e^{x} + 1}) + C

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