`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 \( \int \frac{\sqrt{x}}{1 + 4\sqrt{x^3}} \, dx \), we will use a substitution method. Let's go through the steps systematically.

### Step 1: Substitution
Let \( t = 1 + 4\sqrt{x^3} \).

### Step 2: Differentiate \( t \)
To find \( dt \), we need to differentiate \( t \) with respect to \( x \):
\[ t = 1 + 4\sqrt{x^3} \]
Differentiating both sides:
\[ \frac{dt}{dx} = 4 \cdot \frac{1}{2\sqrt{x^3}} \cdot 3x^2 = \frac{6x^2}{\sqrt{x^3}} = \frac{6x^2}{x^{3/2}} = 6x^{1/2} \]
Thus, we have:
\[ dt = 6\sqrt{x} \, dx \quad \Rightarrow \quad dx = \frac{dt}{6\sqrt{x}} \]

### Step 3: Express \( \sqrt{x} \) in terms of \( t \)
From our substitution:
\[ 4\sqrt{x^3} = t - 1 \quad \Rightarrow \quad \sqrt{x^3} = \frac{t - 1}{4} \quad \Rightarrow \quad \sqrt{x} = \left(\frac{t - 1}{4}\right)^{2/3} \]

### Step 4: Substitute back into the integral
Now substitute \( \sqrt{x} \) and \( dx \) into the integral:
\[ \int \frac{\sqrt{x}}{1 + 4\sqrt{x^3}} \, dx = \int \frac{\left(\frac{t - 1}{4}\right)^{2/3}}{t} \cdot \frac{dt}{6\left(\frac{t - 1}{4}\right)^{1/3}} \]
This simplifies to:
\[ \int \frac{(t - 1)^{2/3}}{6t \cdot 4^{2/3}} \cdot \frac{dt}{(t - 1)^{1/3}} = \int \frac{(t - 1)^{1}}{24t} \, dt \]

### Step 5: Simplify the integral
Now we can simplify the integral:
\[ \int \frac{(t - 1)}{24t} \, dt = \frac{1}{24} \int \left(1 - \frac{1}{t}\right) \, dt = \frac{1}{24} \left(t - \log|t|\right) + C \]

### Step 6: Substitute back \( t \)
Now substitute back \( t = 1 + 4\sqrt{x^3} \):
\[ = \frac{1}{24} \left(1 + 4\sqrt{x^3} - \log|1 + 4\sqrt{x^3}|\right) + C \]

### Final Answer
Thus, the integral \( \int \frac{\sqrt{x}}{1 + 4\sqrt{x^3}} \, dx \) is equal to:
\[ \frac{1 + 4\sqrt{x^3}}{24} - \frac{1}{24} \log|1 + 4\sqrt{x^3}| + C \]