Trig. Indefinite Integral $\int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$

Question

$\displaystyle \int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$

$\underline{\bf{My \; Try}}$:: 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

Answer 1

Mathematica evaluates it as $$\frac{\tan ^{-1}\left(\frac{2 \tan (x)-1}{\sqrt{3}}\right)}{\sqrt{3}}+\frac{1}{6} \log \left(\tan ^2(x)-\tan (x)+1\right)-\frac{1}{3} \log (\tan (x)+1)$$ This can be done via partial fractions. \begin{align*} \int \frac{t}{1 + t^3} \; dt &= \frac 1 3 \int \frac{t+1}{t^2 - t + 1} dt - \frac 1 3 \int \frac{1}{1+t} \; dt\\ &= \frac 1 3 \int \frac{t+1}{ \left ( t - \frac 1 2 \right )^2 + \frac 3 4} \; dt - \frac 1 3 \log(1+t)\\ &= \frac 1 3 \int \frac{t - \frac 1 2 + \frac 3 2 }{ \left ( t - \frac 1 2 \right )^2 + \frac 3 4} \; dt - \frac 1 3 \log(1+t)\\ &= \frac 1 6 \log \left( \left ( t - \frac 1 2 \right )^2 + \frac 3 4 \right ) - \frac 1 3 \log(1+t) + \frac 1 2 \int \frac{1}{ \left ( t - \frac 1 2 \right )^2 + \frac 3 4}\; dt \\ &= \frac 1 6 \log \left( \left ( t - \frac 1 2 \right )^2 + \frac 3 4 \right ) - \frac 1 3 \log(1+t) + \frac 1 {\sqrt{3}} \arctan \left( \frac{t - \frac 1 2 }{ \frac{\sqrt 3}{2 } } \right ) \end{align*}

Answer 2

Good work up to now. But now we do need to use partial fractions; that is the most straightforward approach:

$$\dfrac t{t^3 + 1} = \dfrac t{(t+1)(t^2 - t + 1)} = \dfrac{A}{t+1} + \dfrac {Bt + C}{t^2 - t + 1}$$