int frac{e^{2x}}{sqrt[4]{e^x+1}} dx = | vedclass

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(C) Let $I = \int \frac{e^{2x}}{(e^x+1)^{1/4}} dx$.Substitute $t = (e^x+1)^{1/4}$,then $t^4 = e^x+1$,so $e^x = t^4-1$.Differentiating both sides,$e^x

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$\frac{4}{21}(e^x+1)^{3/4}(3e^x-4)+c$

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(C) Let $I = \int \frac{e^{2x}}{(e^x+1)^{1/4}} dx$.Substitute $t = (e^x+1)^{1/4}$,then $t^4 = e^x+1$,so $e^x = t^4-1$.Differentiating both sides,$e^x dx = 4t^3 dt$.Substituting these into the integral:$I = \int \frac{(t^4-1) \cdot (4t^3 dt)}{t} = 4 \int (t^4-1)t^2 dt$.$I = 4 \int (t^6-t^2) dt = 4 \left( \frac{t^7}{7} - \frac{t^3}{3} \right) + C$.$I = 4t^3 \left( \frac{t^4}{7} - \frac{1}{3} \right) + C$.Substitute $t = (e^x+1)^{1/4}$ back:$I = 4(e^x+1)^{3/4} \left( \frac{e^x+1}{7} - \frac{1}{3} \right) + C$.$I = 4(e^x+1)^{3/4} \left( \frac{3e^x+3-7}{21} \right) + C$.$I = \frac{4}{21}(e^x+1)^{3/4}(3e^x-4) + C$.

$\int \frac{e^{2x}}{\sqrt[4]{e^x+1}} dx =$

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