If I_1 = int frac{e^x}{e^{4x} + e^{2x} + 1} d | vedclass

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(D) For $I_2$, multiply the numerator and denominator by $e^{4x}$:$I_2 = \int \frac{e^{-x} \cdot e^{4x}}{e^{-4x} \cdot e^{4x} + e^{-2x} \cdot e^{4x} +

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$\frac{1}{2} \log \left(\frac{e^x + e^{-x} - 1}{e^x + e^{-x} + 1}\right) + c$

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(D) For $I_2$, multiply the numerator and denominator by $e^{4x}$:$I_2 = \int \frac{e^{-x} \cdot e^{4x}}{e^{-4x} \cdot e^{4x} + e^{-2x} \cdot e^{4x} + 1 \cdot e^{4x}} dx = \int \frac{e^{3x}}{1 + e^{2x} + e^{4x}} dx$.Let $e^x = t$, then $e^x dx = dt$, so $dx = \frac{dt}{t}$.$I_1 = \int \frac{t}{t^4 + t^2 + 1} \cdot \frac{dt}{t} = \int \frac{dt}{t^4 + t^2 + 1}$.$I_2 = \int \frac{t^3}{t^4 + t^2 + 1} \cdot \frac{dt}{t} = \int \frac{t^2}{t^4 + t^2 + 1} dt$.$I_2 - I_1 = \int \frac{t^2 - 1}{t^4 + t^2 + 1} dt = \int \frac{1 - 1/t^2}{t^2 + 1 + 1/t^2} dt = \int \frac{1 - 1/t^2}{(t + 1/t)^2 - 1} dt$.Let $u = t + 1/t$, then $du = (1 - 1/t^2) dt$.$I_2 - I_1 = \int \frac{du}{u^2 - 1} = \frac{1}{2} \log \left| \frac{u - 1}{u + 1} \right| + c$.Substituting $u = e^x + e^{-x}$:$I_2 - I_1 = \frac{1}{2} \log \left| \frac{e^x + e^{-x} - 1}{e^x + e^{-x} + 1} \right| + c$.

If $I_1 = \int \frac{e^x}{e^{4x} + e^{2x} + 1} dx$ and $I_2 = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} dx$, then $I_2 - I_1 =$

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