<s>[INST] Trig. Indefinite Integral $\int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$ $\displaystyle \int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$
$\underline{\bf{My \; Try}}$:: Let $\tan x = t$. Then $\sec^2 xdx = dt\Rightarrow \displaystyle dx = \frac{1}{1+\tan^2 t}dt\Rightarrow dx = \frac{1}{1+t^2}dt$ 
So $\displaystyle \int\frac{t+t^3}{1+t^3}\cdot \frac{1}{1+t^2}dt = \int\frac{t}{1+t^3}dt$
Now My Question is can we solve the Given Integral without Using Partial fraction Method
If Yes How can I solve
plz Help me , Thanks 
 [/INST] Addendum:
Evaluating $ \displaystyle{\int\frac{t}{t^3 + 1}\,\mathrm{d}t} $ without partial fractions:
$$
\begin{aligned}
\int\frac{t}{t^3 + 1}\,\mathrm{d}t&=\frac{1}{2}\int\frac{t+1+t-1}{t^3+1}\,\mathrm{d}t\\
&=\frac{1}{2}\int\frac{t+1}{(t+1)(t^2-t+1)}\,\mathrm{d}t-\frac{1}{2}\int\frac{t^2 - t + 1 - t^2}{t^3 + 1}\,\mathrm{d}t\\
&=\frac{1}{2}\int\frac{\mathrm{d}t}{\left(t-1/2\right)^2 + \left(\sqrt{3}/2\right)^2} - \frac{1}{2}\int\frac{\mathrm{d}t}{t+1} + \frac{1}{6}\int\frac{3t^2}{t^3 + 1}\,\mathrm{d}t\\
&=\frac{1}{\sqrt{3}}\arctan\left(\frac{2t-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left|t+1\right| + \frac{1}{6}\log\left|t^3 + 1\right| + C.
\end{aligned}
$$
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