Mathematics - Indefinite Integration Question with Solution | TestHub

Since, $J=\int \frac{e^{3x}}{1+e^{2x}+e^{4x}}dx$ $$\begin{array}{rl}\therefore J-I& =\int \frac{\left(e^{3x}-e^x\right)}{1+e^{2x}+e^{4x}}dx=\int \frac{\left(u^2-1\right)}{1+u^2+u^4}du\\ & =\int \frac{\left(1-\frac{1}{u^2}\right)}{1+\frac{1}{u^2}+u^2}du=\int \frac{\left(1-\frac{1}{u^2}\right)}{{\left(u+\frac{1}{u}\right)}^2-1}du\\ & =\int \frac{dt}{t^2-1}\\ & =\frac{1}{2}\log \left|\frac{t-1}{t+1}\right|+C\\ & =\frac{1}{2}\log \left|\frac{u^2-u+1}{u^2+u+1}\right|+C=\frac{1}{2}\log \left|\frac{e^{2x}-e^x+1}{e^{2x}+e^x+1}\right|+C\end{array}$$