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Evaluate the following integral:

$$\int\sqrt[3]{x^3+\frac 1 {x^3}} dx$$

For $x^3=u$ and $1+\frac 1 {u^2}=v$, getting $$-\frac 1 6 \int(v-1)^{-\frac 4 3}v^{\frac 1 3} dv$$, what is the next step?

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  • $\begingroup$ This can be expressed as Gauss Hypergeometric series. $\endgroup$
    – Z Ahmed
    Feb 7, 2021 at 5:13

2 Answers 2

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I would chose a different substitution:

$$I =\int \big(x^3 +\frac{1}{x^3} \big)^{\frac{1}{3}}dx$$ $$=\int \frac{(x^6 +1)^{\frac{1}{3}}}{x} dx $$ Let: $ u=(x^6+1)^{\frac{1}{3}} $, then: $x^6=u^3-1 \text{ , and } dx= \frac{(x^6 +1)^{\frac{2}{3}}}{2x^5} du$ . $$I=\frac{1}{2} \int \frac{u^3}{u^3-1}du$$ $$=\frac{1}{2} \int \frac{1}{u^3-1}du \text{ }+ \text{ } \frac{1}{2}u$$

The first integral can easily be solved now with partial fraction decomposition; and using the derivative of the arctan function.

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  • $\begingroup$ @ZAhmed I don't think the solution is incorrect; in fact it matches the other solution that just got posted. Using my substitution the integrand becomes: $ (x^3 + \frac{1}{x^3})^{1/3} dx= \frac{(x^6 +1)^{1/3}}{x} dx = \frac{(x^6 +1)^{1/3}}{x} \frac{(x^6 +1)^{2/3}}{2x^5} du = \frac{1}{2} \frac{((x^6 +1)^{1/3})^3 }{x^6} du = \frac{1}{2} \frac{u^3}{u^3-1}$ $\endgroup$ Feb 7, 2021 at 5:50
  • $\begingroup$ Sorry you are right. $\endgroup$
    – Z Ahmed
    Feb 7, 2021 at 6:57
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Continue with

$$I=-\frac 1 6 \int(v-1)^{-\frac 4 3}v^{\frac 1 3} dv = -\frac 1 6 \int \frac1{v-1} \sqrt[3]{\frac v{v-1} } dv $$

and let $t^3 = \frac v{v-1}$. Then, the integral becomes \begin{align} I &= \frac12 \int \frac{t^3}{t^3-1}dt = \frac12 t +\frac16\ln \frac {t-1}{\sqrt{t^2+t+1}}-\frac1{2\sqrt3}\tan^{-1}\frac{2t+1}{\sqrt3} \end{align}

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