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Is there a simpler way to do it? Please, no hyperbolic substitutions.

$$\int x^\frac{1}{2}(1+2x^\frac{1}{3})^{-{3}}$$

$ p = -3 $ and because p is an negative integer, we use $ x = t^6 $ supstitution, because six is the smallest denominator (please, someone edit this, I don't know the English word for it, I hope that you get what I mean by this) of $m=\frac{1}{2},n=-\frac{1}{3}$

$$ dx = 6t^5dt$$

$$ 6 \int \sqrt{t^6}(1+2t^2)^{-3}t^5dt$$ $$6 \int|t^3|\frac{t^5dt}{(1+2t^2)^{3}}$$

For $t\geq0$

$$6 \int\frac{t^8dt}{(1+2t^2)^{3}}$$

Another supstitution, $\sqrt{2}t=k, dt = \frac{dk}{\sqrt{2}}$

$$\frac{6}{|16|\sqrt{2}}\int\frac{k^8dk}{{(1+k^2)}^3}$$

$k=\tan\theta, dk = \sec^2\theta d\theta$

$$\frac{6}{|16|\sqrt{2}}\int\frac{\tan^8\theta \sec^2\theta d\theta}{\sec^6\theta}$$

$$\frac{6}{|16|\sqrt{2}}\int(\frac{(\sec^2 \theta-1)}{\sec\theta})^4 d\theta = \frac{6}{|16|\sqrt{2}}\int(\sec \theta - \frac{(1)}{\sec\theta})^4 d\theta $$

$$\frac{6}{|16|\sqrt{2}}\int(\sec \theta - \frac{1}{\sec\theta})^4 d\theta = \frac{6}{|16|\sqrt{2}}\int(\frac{1}{\cos \theta} - \cos \theta)^4 d\theta $$

$$\frac{6}{|16|\sqrt{2}}\int(\frac{1-\cos^2\theta}{\cos \theta})^4 d\theta$$

THEN we just expand the binomial, plug back in the substitutions Simple... But it is kind of bothersome to have so many substitutions, and to return from every substitution. Especially, when it is not allowed to use the principal square root.

Could you please, point me to some good literature on integration of irrational functions?

EDIT: I made a serious mistake in the process of doing the integral. Thank you for pointing it out. Can you, please, answer my questions. Thank you.

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  • $\begingroup$ i think this is wring, since $$x^{-1/3}=\frac{1}{x^{1/3}}$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 23 '17 at 16:17
  • $\begingroup$ @Dr.SonnhardGraubner Mistake in the beginning, sorry. $\endgroup$ – Shocky2 Jul 23 '17 at 16:22
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HINT: simplify your integrand to $$\int \frac{x^{1/2}}{1+6x^{1/3}+12x^{2/3}+8x}$$ and set $$x=u^6 , dx = 6u^5$$

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