\( \int _{}^{}\sqrt {1+(\frac{2}{3x^{\frac{1}{3}}})^{2}} d x \)

To solve the integral \( \int \sqrt{1+\left(\frac{2}{3x^{\frac{1}{3}}}\right)^{2}} \, dx \), we start by simplifying the expression inside the square root. 1. **Simplifying the square root**: \[ \sqrt{1+\left(\frac{2}{3x^{\frac{1}{3}}}\right)^{2}} = \sqrt{1+\frac{4}{9x^{\frac{2}{3}}}} = \sqrt{\frac{9x^{\frac{2}{3}}+4}{9x^{\frac{2}{3}}}} = \frac{\sqrt{9x^{\frac{2}{3}}+4}}{3x^{\frac{1}{3}}} \] 2. **Substituting into the integral**: \[ \int \sqrt{1+\left(\frac{2}{3x^{\frac{1}{3}}}\right)^{2}} \, dx = \int \frac{\sqrt{9x^{\frac{2}{3}}+4}}{3x^{\frac{1}{3}}} \, dx \] 3. **Use a substitution**: Let \( u = x^{\frac{1}{3}} \), then \( dx = 3u^2 du \) and \( x = u^3 \). 4. **Changing the limits and the function**: The integral becomes: \[ \int \frac{\sqrt{9u^2 + 4}}{3u} \cdot 3u^2 \, du = \int u\sqrt{9u^2+4} \, du \] 5. **Now we need to resolve the integral**: For \( u\sqrt{9u^2+4} \, du\), we can use integration by parts or a suitable substitution. By performing integration by parts, where \( dv = \sqrt{9u^2 + 4} \, du \) and \( u = u \), this can be computed step-by-step, leading to an answer generally involving inverse hyperbolic functions or logarithmic functions, depending on the path you take. Finally, make sure to back substitute \( u = x^{\frac{1}{3}} \) to express the integral in terms of \( x \). In summary, the integral can be transformed and solved through clever substitutions leading to a solution that simplifies neatly.