Explanation

(c)
Since, $J = \int_{}^{}\frac{e^{3x}}{1 + e^{2x} + e^{4x}}dx$
$\therefore\ \ J - I = \int_{}^{}\frac{(e^{3x} - e^{x})}{1 + e^{2x} + e^{4x}}\ dx$
$= \int_{}^{}\frac{(u^{2} - 1)}{1 + u^{2} + u^{4}}du$ $\lbrack u = e^{x}\rbrack$
$= \int_{}^{}\frac{\left( 1 - \frac{1}{u^{2}} \right)}{1 + \frac{1}{u^{2}} + u^{2}}\ du = \int_{}^{}\frac{\left( 1 - \frac{1}{u^{2}} \right)}{\left( u + \frac{1}{u} \right)^{2} - 1}\ du$
$= \int_{}^{}\frac{dt}{t^{2} - 1}$ $\left\lbrack put\ u + \frac{1}{u} = t \Rightarrow \left( 1 - \frac{1}{u^{2}} \right)du = dt \right\rbrack$
$= \frac{1}{2}\log\left| \frac{t - 1}{t + 1} \right| + c = \frac{1}{2}\log\left| \frac{u^{2} - u + 1}{u^{2} + u + 1} \right| + c$
$= \frac{1}{2}\log\left| \frac{e^{2x} - e^{x} + 1}{e^{2x} + e^{x} + 1} \right| + c$