IIT-JEE 2008 · Paper 2, Mathematics question 8

Rewrite J. Multiplying through by $e^{4x}$: $J=\displaystyle\int\frac{e^{3x}}{e^{4x}+e^{2x}+1}dx$.

Difference. With $u=e^x$:
$$J-I=\int\frac{u^2-1}{u^4+u^2+1}du=\int\frac{1-\frac1{u^2}}{\left(u+\frac1u\right)^2-1}du.$$
Put $v=u+\frac1u$:
$$\int\frac{dv}{v^2-1}=\frac12\log\frac{v-1}{v+1}=\frac12\log\frac{u^2-u+1}{u^2+u+1}.$$

(C) $\frac12\log\left(\dfrac{e^{2x}-e^x+1}{e^{2x}+e^x+1}\right)+C$