`int root 3 (x) root 7 (1+ root3 (x ^(4)))` dx is equal to

Options

A. `(21)/(32) (1+ root 3 (x ^(4))) ^(8//7)+C`
B. `(32)/(21) (1 + root 3 ( x ^(4))) ^(8//7) +C`
C. `(7)/(32) (1 + root 3 ( x ^(4))) ^(8//7) +C`
D. None of these

Answer

To solve the integral \( I = \int x^{\frac{1}{3}} \sqrt[7]{1 + \sqrt[3]{x^4}} \, dx \), we will follow these steps:

### Step 1: Substitute \( y = \sqrt[3]{x} \)

Let \( y = \sqrt[3]{x} \). Then, we have:
\[ x = y^3 \quad \text{and} \quad dx = 3y^2 \, dy \]

### Step 2: Rewrite the integral in terms of \( y \)

Substituting \( x \) and \( dx \) into the integral, we get:
\[ I = \int (y^3)^{\frac{1}{3}} \sqrt[7]{1 + \sqrt[3]{(y^3)^4}} \cdot 3y^2 \, dy \]
This simplifies to:
\[ I = \int y \sqrt[7]{1 + y^4} \cdot 3y^2 \, dy = 3 \int y^3 \sqrt[7]{1 + y^4} \, dy \]

### Step 3: Simplify the integral

Now, we can write:
\[ I = 3 \int y^3 (1 + y^4)^{\frac{1}{7}} \, dy \]

### Step 4: Use substitution for integration

Let \( u = 1 + y^4 \). Then, we differentiate:
\[ du = 4y^3 \, dy \quad \Rightarrow \quad dy = \frac{du}{4y^3} \]
Substituting \( y^3 \) in terms of \( u \):
\[ I = 3 \int y^3 u^{\frac{1}{7}} \cdot \frac{du}{4y^3} = \frac{3}{4} \int u^{\frac{1}{7}} \, du \]

### Step 5: Integrate with respect to \( u \)

Now, we can integrate:
\[ \int u^{\frac{1}{7}} \, du = \frac{u^{\frac{1}{7} + 1}}{\frac{1}{7} + 1} + C = \frac{u^{\frac{8}{7}}}{\frac{8}{7}} + C = \frac{7}{8} u^{\frac{8}{7}} + C \]
Thus,
\[ I = \frac{3}{4} \cdot \frac{7}{8} u^{\frac{8}{7}} + C = \frac{21}{32} u^{\frac{8}{7}} + C \]

### Step 6: Substitute back for \( u \)

Recall that \( u = 1 + y^4 \) and \( y = \sqrt[3]{x} \):
\[ I = \frac{21}{32} (1 + (\sqrt[3]{x})^4)^{\frac{8}{7}} + C = \frac{21}{32} (1 + x^{\frac{4}{3}})^{\frac{8}{7}} + C \]

### Final Answer

Thus, the integral evaluates to:
\[ I = \frac{21}{32} (1 + x^{\frac{4}{3}})^{\frac{8}{7}} + C \]