If \(\mathrm{I}=\int \frac{\mathrm{e}^x}{\mathrm{e}^{4 x}+\mathrm{e}^{2 x}+1} \mathrm{~d} x\) and \(\mathrm{J}=\int \frac{\mathrm{e}^{-x}}{\mathrm{e}^{-4 x}+\mathrm{e}^{-2 x}+1} \mathrm{~d} x\) then for any arbitrary constant c, the value of \(\mathrm{J}-\mathrm{I}\) equals

\(\begin{aligned} \mathrm{J}-\mathrm{I} & =\int\left(\frac{\mathrm{e}^{-x}}{\mathrm{e}^{-4 x}+\mathrm{e}^{-2 x}+1}-\frac{\mathrm{e}^x}{\mathrm{e}^{4 x}+\mathrm{e}^{2 x}+1}\right) \mathrm{d} x \\ & =\int\left(\frac{\mathrm{e}^{3 x}}{\mathrm{e}^{4 x}+\mathrm{e}^{2 x}+1}-\frac{\mathrm{e}^x}{\mathrm{e}^{4 x}+\mathrm{e}^{2 x}+1}\right) \mathrm{d} x \\ & =\int \frac{\left(\mathrm{e}^{2 x}-1\right) \mathrm{e}^x}{\mathrm{e}^{4 x}+\mathrm{e}^{2 x}+1} \mathrm{~d} x\end{aligned}\)
Put \(\mathrm{e}^x=\mathrm{t} \Rightarrow \mathrm{e}^x \mathrm{~d} x=\mathrm{dt}\)
\(\therefore \quad \mathrm{J}-\mathrm{I}=\int \frac{\mathrm{t}^2-1}{\mathrm{t}^4+\mathrm{t}^2+1} \mathrm{dt}=\int \frac{1-\frac{1}{\mathrm{t}^2}}{\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^2-1} \mathrm{dt}\)
Put \(\mathrm{t}+\frac{1}{\mathrm{t}}=y\)
\(\Rightarrow\left(1-\frac{1}{\mathrm{t}^2}\right) \mathrm{dt}=\mathrm{d} y\)
\(\therefore \quad \mathrm{J}-\mathrm{I}=\int \frac{\mathrm{d} y}{y^2-1^2}=\frac{1}{2} \log \left|\frac{y-1}{y+1}\right|+\mathrm{c}\)
\(\begin{aligned} & =\frac{1}{2} \log \left|\frac{\mathrm{t}+\frac{1}{\mathrm{t}}-1}{\mathrm{t}+\frac{1}{\mathrm{t}}+1}\right|+\mathrm{c} \\ & =\frac{1}{2} \log \left|\frac{\mathrm{t}^2-\mathrm{t}+1}{\mathrm{t}^2+\mathrm{t}+1}\right|+\mathrm{c} \\ & =\frac{1}{2} \log \left|\frac{\mathrm{e}^{2 x}-\mathrm{e}^x+1}{\mathrm{e}^{2 x}+\mathrm{e}^x+1}\right|+\mathrm{c}\end{aligned}\)