$\int \frac{x+1}{\sqrt[3]{x^{2}}} d x$

Step 1 ：Let's rewrite the integrand as \(\frac{x+1}{\sqrt[3]{x^{2}}} = \frac{x+1}{x^{\frac{2}{3}}} = x^{\frac{1}{3}} + x^{-\frac{2}{3}}\)

Step 2 ：Now, we can integrate term by term:

Step 3 ：\(\int (x^{\frac{1}{3}} + x^{-\frac{2}{3}}) dx = \int x^{\frac{1}{3}} dx + \int x^{-\frac{2}{3}} dx\)

Step 4 ：Using the power rule for integration, we have:

Step 5 ：\(\int x^{\frac{1}{3}} dx = \frac{3}{4} x^{\frac{4}{3}} + C_1\) and \(\int x^{-\frac{2}{3}} dx = -\frac{3}{x^{\frac{1}{3}}} + C_2\)

Step 6 ：So, the final integral is:

Step 7 ：\(\int \frac{x+1}{\sqrt[3]{x^{2}}} dx = \frac{3}{4} x^{\frac{4}{3}} - \frac{3}{x^{\frac{1}{3}}} + C\)

Step 8 ：Where \(C = C_1 + C_2\) is the constant of integration.

Step 9 ：Thus, the final answer is \(\boxed{\frac{3}{4} x^{\frac{4}{3}} - \frac{3}{x^{\frac{1}{3}}} + C}\)