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 
$$\int\frac{t}{t^3 + 1}dt.$$
 A: 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.
A: Hint: $\frac{t}{t^{3}+1} = \frac{A}{t+1} + \frac{Bt + C}{t^{2}-t+1}$ where $A$ and $B$ and $C$ are constants to be found
Can you solve it now?
In case you get stuck:
$A = -\frac{1}{3}$
$B = \frac{1}{3}$
$C = \frac{1}{3}$
Then we get $I = \int -\frac{1}{3(t+1)} + \frac{t}{3(t^{2}-t+1)} + \frac{1}{3(t^{2}-t+1)}dt$
A: $\textbf{Hint.}$ Firstly, 
$$f(t)=\frac{t}{t^3+1}=\frac{t}{(t+1)(t^2-t+1)}=\frac{t+1}{3(t^2-t+1)}-\frac{1}{3(t+1)}$$
The second integral is immediate (a logarithmic function). The first needs more work, but it can be reduced to the integral of a logaritmic plus an $\arctan$
A: 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

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

