Integration of a binomial differential
`I=intx^(5)(1+x^(3))^((2)/(3))dx.`

Answer

To solve the integral \( I = \int x^5 (1 + x^3)^{\frac{2}{3}} \, dx \), we will use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = 1 + x^3 \). Then, we differentiate \( t \) with respect to \( x \):
\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
We also need to express \( x^5 \) in terms of \( t \). Since \( t = 1 + x^3 \), we have:
\[ x^3 = t - 1 \quad \Rightarrow \quad x = (t - 1)^{\frac{1}{3}} \]
Thus,
\[ x^5 = (x^3)^{\frac{5}{3}} = (t - 1)^{\frac{5}{3}} \]

### Step 2: Rewrite the Integral
Substituting \( t \) and \( dx \) into the integral, we get:
\[ I = \int (t - 1)^{\frac{5}{3}} t^{\frac{2}{3}} \cdot \frac{dt}{3x^2} \]
From \( t = 1 + x^3 \), we can express \( x^2 \) as:
\[ x^2 = (t - 1)^{\frac{2}{3}} \]
Thus, the integral becomes:
\[ I = \int (t - 1)^{\frac{5}{3}} t^{\frac{2}{3}} \cdot \frac{dt}{3(t - 1)^{\frac{2}{3}}} \]
This simplifies to:
\[ I = \frac{1}{3} \int (t - 1)^{\frac{5}{3} - \frac{2}{3}} t^{\frac{2}{3}} \, dt = \frac{1}{3} \int (t - 1)^{1} t^{\frac{2}{3}} \, dt \]

### Step 3: Expand and Integrate
Now we expand the integrand:
\[ (t - 1) t^{\frac{2}{3}} = t^{\frac{5}{3}} - t^{\frac{2}{3}} \]
Thus, the integral becomes:
\[ I = \frac{1}{3} \left( \int t^{\frac{5}{3}} \, dt - \int t^{\frac{2}{3}} \, dt \right) \]

### Step 4: Perform the Integrals
Now we can integrate each term:
\[ \int t^{\frac{5}{3}} \, dt = \frac{t^{\frac{5}{3} + 1}}{\frac{5}{3} + 1} = \frac{t^{\frac{8}{3}}}{\frac{8}{3}} = \frac{3}{8} t^{\frac{8}{3}} \]
\[ \int t^{\frac{2}{3}} \, dt = \frac{t^{\frac{2}{3} + 1}}{\frac{2}{3} + 1} = \frac{t^{\frac{5}{3}}}{\frac{5}{3}} = \frac{3}{5} t^{\frac{5}{3}} \]

### Step 5: Combine Results
Substituting back into \( I \):
\[ I = \frac{1}{3} \left( \frac{3}{8} t^{\frac{8}{3}} - \frac{3}{5} t^{\frac{5}{3}} \right) + C \]
\[ I = \frac{1}{8} t^{\frac{8}{3}} - \frac{1}{5} t^{\frac{5}{3}} + C \]

### Step 6: Substitute Back for \( t \)
Recall that \( t = 1 + x^3 \):
\[ I = \frac{1}{8} (1 + x^3)^{\frac{8}{3}} - \frac{1}{5} (1 + x^3)^{\frac{5}{3}} + C \]

### Final Answer
Thus, the final result for the integral is:
\[ I = \frac{1}{8} (1 + x^3)^{\frac{8}{3}} - \frac{1}{5} (1 + x^3)^{\frac{5}{3}} + C \]