$\displaystyle \int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$
$\underline{\bf{My \; Try}}$:: 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
Best Answer
Good work up to now. But now we do need to use partial fractions; that is the most straightforward approach:
$$\dfrac t{t^3 + 1} = \dfrac t{(t+1)(t^2 - t + 1)} = \dfrac{A}{t+1} + \dfrac {Bt + C}{t^2 - t + 1}$$