Let I=int frac{e^x}{e^{4 x}+e^{2 x}+1} d x an | vedclass

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(C) Given $J = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} dx$. Multiplying numerator and denominator by $e^{4x}$,we get $J = \int \frac{e^{3x}}{1 + e^{

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

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(C) Given $J = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} dx$. Multiplying numerator and denominator by $e^{4x}$,we get $J = \int \frac{e^{3x}}{1 + e^{2x} + e^{4x}} dx$.Now,$J - I = \int \frac{e^{3x} - e^x}{e^{4x} + e^{2x} + 1} dx = \int \frac{e^x(e^{2x} - 1)}{e^{4x} + e^{2x} + 1} dx$.Let $z = e^x$,then $dz = e^x dx$. The integral becomes $\int \frac{z^2 - 1}{z^4 + z^2 + 1} dz$.Divide numerator and denominator by $z^2$: $\int \frac{1 - 1/z^2}{z^2 + 1 + 1/z^2} dz = \int \frac{1 - 1/z^2}{(z + 1/z)^2 - 1} dz$.Let $u = z + 1/z$,then $du = (1 - 1/z^2) dz$. The integral becomes $\int \frac{du}{u^2 - 1} = \frac{1}{2} \ln \left| \frac{u - 1}{u + 1} \right| + C$.Substituting $u = e^x + e^{-x}$ back: $\frac{1}{2} \ln \left| \frac{e^x + e^{-x} - 1}{e^x + e^{-x} + 1} \right| + C = \frac{1}{2} \ln \left( \frac{e^{2x} - e^x + 1}{e^{2x} + e^x + 1} \right) + C$.

Let $I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x$ and $J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x$. Then,for an arbitrary constant $C$,the value of $J-I$ equals

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