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

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Aicharya · Accepted Answer

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

If $\mathrm{I}=\int \frac{\mathrm{e}^x}{\mathrm{e}^{4 x}+\mathrm{e}^{2 x}+1} \mathrm{~d} x$ and…

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