`int(root(3)(1+root(4)(x)))/(sqrtx)\ dx` is equal to :

Options

A. `12(((1+root(4)(x))^(7//3))/(7)+((1+root(4)(x))^(4//3))/(4))+C`
B. `12(((1+root(4)(x))^(7//3))/(7)-((1+root(4)(x))^(4//3))/(4))+C`
C. `6(((1+root(4)(x))^(7//3))/(7)-((1+root(4)(x))^(4//2))/(4))+C`
D. None of these

Answer

To solve the integral

\[ I = \int \frac{\sqrt[3]{1 + \sqrt[4]{x}}}{\sqrt{x}} \, dx, \]

we will use substitution to simplify the expression.

### Step 1: Substitution
Let

\[ t = \sqrt[4]{x} \implies x = t^4. \]

Then, we can find \(dx\):

\[ dx = 4t^3 \, dt. \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral in terms of \(t\):

\[ \sqrt{x} = \sqrt{t^4} = t^2, \]

and

\[ \sqrt[4]{x} = t. \]

Substituting these into the integral gives:

\[ I = \int \frac{\sqrt[3]{1 + t}}{t^2} (4t^3) \, dt = 4 \int \sqrt[3]{1 + t} \cdot t \, dt. \]

### Step 3: Simplify the Integral
Now we simplify the integral:

\[ I = 4 \int t \sqrt[3]{1 + t} \, dt. \]

### Step 4: Further Substitution
Let

\[ u = 1 + t \implies du = dt \quad \text{and} \quad t = u - 1. \]

Now, substituting \(t\) in terms of \(u\):

\[ I = 4 \int (u - 1) \sqrt[3]{u} \, du = 4 \int (u^{4/3} - u^{1/3}) \, du. \]

### Step 5: Integrate
Now we can integrate term by term:

\[ I = 4 \left( \frac{u^{7/3}}{7/3} - \frac{u^{4/3}}{4/3} \right) + C = 4 \left( \frac{3}{7} u^{7/3} - \frac{3}{4} u^{4/3} \right) + C. \]

### Step 6: Substitute Back
Now we substitute back \(u = 1 + t\):

\[ I = \frac{12}{7} (1 + t)^{7/3} - 3 (1 + t)^{4/3} + C. \]

Finally, substituting back \(t = \sqrt[4]{x}\):

\[ I = \frac{12}{7} \left(1 + \sqrt[4]{x}\right)^{7/3} - 3 \left(1 + \sqrt[4]{x}\right)^{4/3} + C. \]

### Final Answer
Thus, the integral

\[ \int \frac{\sqrt[3]{1 + \sqrt[4]{x}}}{\sqrt{x}} \, dx = \frac{12}{7} \left(1 + \sqrt[4]{x}\right)^{7/3} - 3 \left(1 + \sqrt[4]{x}\right)^{4/3} + C. \]

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