Perform this integration: $\int\frac{t}{t^3 + 1}$ I am solving the following question
$$\int\frac{\sin x}{\sin^{3}x + \cos^{3}x}dx.$$
I have been able to reduce it to the following form by diving numerator and denominator by $\cos^{3}x$ and then substituting $\tan x$ for $t$ and am getting the following equation. Should Iis there any other way use partial fraction to integrate it further or 
$$\int\frac{t}{t^3 + 1}dt.$$

We have that $-1,\omega,\omega^{-1}$ are the roots of $t^3+1$. In particular, $\frac{t}{t^3+1}$ can be represented as
$$ \frac{t}{t^3+1} = \frac{A}{t+1}+\frac{B}{t-\omega}+\frac{C}{t-\omega^{-1}}$$
with $A+B+C=0$ and 
$$ A = \lim_{t\to -1}\frac{t(t+1)}{t^3+1} = \lim_{t\to -1}\frac{t}{t^2-t+1} = -\frac{1}{3} $$
hence:
$$ \frac{t}{t^3+1} = -\frac{1}{3}\cdot \frac{1}{t+1}+\frac{1}{3}\cdot\frac{t+1}{t^2-t+1} $$
and

$$ \int\frac{t}{t^3+1}\,dt = C+\frac{1}{\sqrt{3}}\,\arctan\left(\frac{2t-1}{\sqrt{3}}\right)+\frac{1}{3}\,\log(1+t)-\frac{1}{6}\,\log(1-t+t^2). $$

