`int(sqrt(x))/(1+4sqrt(x^(3)))dx` is equal to

Options

A. `(4)/(3)[1+x^(3//4)+log_(e)(1+x^(3//4))]+C`
B. `(4)/(3)[1+x^(3//4)-log_(e)(1+x^(3//4))]+C`
C. `(4)/(3)[1+x^(3//4)+log_(e)(1+x^(3//4))]+C`
D. none of these

Answer

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

### Step 1: Substitution
Let \( T = 4\sqrt{x^3} \). Then we differentiate \( T \) with respect to \( x \):
\[ T = 4x^{3/2} \]
Differentiating both sides gives:
\[ \frac{dT}{dx} = 4 \cdot \frac{3}{2} x^{1/2} = 6\sqrt{x} \]
Thus, we can express \( \sqrt{x} \, dx \) in terms of \( dT \):
\[ \sqrt{x} \, dx = \frac{dT}{6} \]

### Step 2: Rewrite the Integral
Substituting \( T \) into the integral, we have:
\[ I = \int \frac{\frac{dT}{6}}{1 + T} \]
This simplifies to:
\[ I = \frac{1}{6} \int \frac{dT}{1 + T} \]

### Step 3: Integrate
The integral \( \int \frac{dT}{1 + T} \) is a standard integral:
\[ \int \frac{dT}{1 + T} = \log(1 + T) + C \]
Thus, substituting back, we get:
\[ I = \frac{1}{6} \left( \log(1 + T) + C \right) \]

### Step 4: Substitute Back for \( T \)
Now we substitute back \( T = 4\sqrt{x^3} \):
\[ I = \frac{1}{6} \log(1 + 4\sqrt{x^3}) + C \]

### Final Answer
Thus, the final result for the integral is:
\[ I = \frac{1}{6} \log(1 + 4\sqrt{x^3}) + C \]

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