$\int {e^{3x} - e^x \over e^{4x} + e^{2x} + 1} dx$ 
$$I = \int {e^{3x} - e^x \over e^{4x} + e^{2x} + 1} dx$$

Substituting for $e^x$, 
$$I = \int {u^2 - 1 \over u^4 + u^2 + 1} du = \int { u^4 + u^2 + 1 + - 2 - u^4 \over u^4 + u^2 + 1} du = u - \int {u^4 + 2 \over u^4 + u^2 + 1} du $$
Now I don't know anything I can do to last integral except partial fraction decomposition but I am pretty sure that $u^4 + u^2 + 1$ does not have any factors in real numbers.
Is this integral computable on real numbers ? How do I compute it ?
 A: Observe that $$u^4 + u^2 + 1 = (u^4 + 2u^2 + 1) - u^2 = (u^2 + 1)^2 - u^2 = (u^2 + u + 1)(u^2 - u + 1).$$  From this, we try a partial fraction decomposition of the form $$\frac{Au + B}{u^2 - u + 1} + \frac{Cu + D}{u^2 + u + 1} = \frac{u^2 - 1}{u^4 + u^2 + 1},$$ and after multiplying out and comparing like coefficients of $u$, we have $$\begin{align*} A + C &= 0 \\ A+B-C+D &= 1 \\
A+B+C-D &= 0 \\
B+D &= -1. \end{align*}$$  From here, we get $$B = D = -1/2,$$ and $$A = 1, C = -1.$$  Consequently, the integrand becomes $$\frac{1}{2}\left( \frac{2u-1}{u^2-u+1} - \frac{2u+1}{u^2+u+1}\right),$$ and the rest is straightforward.  As a bonus, you even find that the numerator of each term is the derivative of the respective denominator.  You can't get much nicer than that.
A: We don't actually need tedious Partial Fraction Decomposition as
$$\dfrac{u^2-1}{u^4+Au^2+1}=\dfrac{1-\dfrac1{u^2}}{u^2+A+\dfrac1{u^2}}$$ where $A$ is an arbitrary constant.
Now as $\displaystyle\int\left(1-\dfrac1{u^2}\right)du=u+\dfrac1u,$
write $\displaystyle u^2+A+\dfrac1{u^2}=\left(u+\dfrac1u\right)^2+A-2$ and set $u+\dfrac1u=v$
