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

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

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  • $\begingroup$ The Maple command $$Student[Calculus1]:-IntTutor((tan(x)+tan(x)^3)/(1+tan(x)^3), x) $$ outputs $$1/6\,\ln \left( \left( \tan \left( x \right) \right) ^{2}-\tan \left( x \right) +1 \right) +1/3\,\sqrt {3}\arctan \left( 1/3\,\sqrt {3} \left( 2\,\tan \left( x \right) -1 \right) \right) -1/3\,\ln \left( \tan \left( x \right) +1 \right) $$ step by step with explanations. $\endgroup$ – user64494 Sep 24 '13 at 20:36
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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*}

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  • $\begingroup$ @amWhy after replacing $t = \tan(x)$ I get $$ \frac 1 6 \log( \tan(x)^2 - \tan(x) + 1) - \frac1 3 \log(1+\tan(x)) + \frac 1{\sqrt 3 } \arctan( (2 \tan(x)- 1)/\sqrt 3 )$$ I think i should get eye check up. could you point out where it is not matching? $\endgroup$ – Santosh Linkha Sep 24 '13 at 15:35
  • $\begingroup$ No, it matches. Just multiply by $2$ inside the $\arctan$ term to get $2t-1\over \sqrt 3$ inside. $\endgroup$ – abiessu Sep 24 '13 at 15:38
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Good work up to now. But now we do need to use partial fractions; that is the most straightforward approach:

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

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  • $\begingroup$ Let me know if you can take it from here. $\endgroup$ – Namaste Sep 24 '13 at 15:14
  • $\begingroup$ Your coefficients should be $A = - \frac 13$, $B = C = \frac 13$ $\endgroup$ – Namaste Sep 24 '13 at 15:23
  • $\begingroup$ very clean answer +1 $\endgroup$ – Amzoti Sep 25 '13 at 0:33
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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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  • $\begingroup$ You used what essentially amounts to partial fractions in the second and third lines, though... $\endgroup$ – user111187 Jan 9 '15 at 11:11

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