<s>[INST] 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 
$$\int\frac{t}{t^3 + 1}dt.$$
 [/INST] Hint:
$\int \frac{t}{t^3+1} dx = \int \frac{t+t^2-t^2}{t^3+1} dx =\int \frac{t(t+1)}{t^3+1} dx-\frac{1}{3}\int \frac{3t^2}{t^3+1} dx=\int \frac{t(t+1)}{(t+1)(t^2-t+1)} dx-\frac{1}{3}\ln(t^3+1)$
$=\frac{1}{2}\int \frac{2t}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)=\frac{1}{2}\int \frac{2t-1+1}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)$
$=\frac{1}{2}\int \frac{2t-1}{t^2-t+1} dx+\frac{1}{2}\int \frac{1}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)$
$=\frac{1}{2}\ln(t^2-t+1)+\frac{1}{2}\int \frac{1}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)$
The integral can be evaluated by partial fractions with complex roots.
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