`int(x+3sqrt(x^(2))+6sqrt(x))/(x(1+3sqrt(x)))dx`

Options

A. `(3)/(2)x^(2//3)+6tan^(-1)x^(1//6)+C`
B. `(3)/(2)x^(2//3)-6tan^(-1)x^(1//6)+C`
C. `-(3)/(2)x^(2//3)-6tan^(-1)x^(1//6)+C`
D. none of these

Answer

To solve the integral

\[ \int \frac{x + 3\sqrt{x^2} + 6\sqrt{x}}{x(1 + 3\sqrt{x})} \, dx, \]

we will simplify the expression step by step.

### Step 1: Simplify the integrand

First, we rewrite the integrand:

\[ \int \frac{x + 3\sqrt{x^2} + 6\sqrt{x}}{x(1 + 3\sqrt{x})} \, dx. \]

Notice that \(\sqrt{x^2} = x\) for \(x \geq 0\). Thus, we can rewrite the numerator:

\[ x + 3\sqrt{x^2} + 6\sqrt{x} = x + 3x + 6\sqrt{x} = 4x + 6\sqrt{x}. \]

Now, substituting this back into the integral gives us:

\[ \int \frac{4x + 6\sqrt{x}}{x(1 + 3\sqrt{x})} \, dx. \]

### Step 2: Split the integral

We can split the integral into two parts:

\[ \int \frac{4x}{x(1 + 3\sqrt{x})} \, dx + \int \frac{6\sqrt{x}}{x(1 + 3\sqrt{x})} \, dx. \]

This simplifies to:

\[ \int \frac{4}{1 + 3\sqrt{x}} \, dx + \int \frac{6\sqrt{x}}{x(1 + 3\sqrt{x})} \, dx. \]

### Step 3: Simplify the second integral

For the second integral, we can rewrite \(\frac{6\sqrt{x}}{x} = \frac{6}{\sqrt{x}}\):

\[ \int \frac{6}{\sqrt{x}(1 + 3\sqrt{x})} \, dx. \]

### Step 4: Substitution for the first integral

For the first integral, let \(u = 1 + 3\sqrt{x}\). Then, we have:

\[ \sqrt{x} = \frac{u - 1}{3} \quad \text{and} \quad x = \left(\frac{u - 1}{3}\right)^2. \]

Now, we differentiate \(u\):

\[ du = \frac{3}{2\sqrt{x}} \, dx \quad \Rightarrow \quad dx = \frac{2\sqrt{x}}{3} \, du = \frac{2(u - 1)}{9} \, du. \]

Substituting this into the integral gives:

\[ \int \frac{4}{u} \cdot \frac{2(u - 1)}{9} \, du = \frac{8}{9} \int \frac{u - 1}{u} \, du = \frac{8}{9} \left( \int 1 \, du - \int \frac{1}{u} \, du \right). \]

### Step 5: Solve the integrals

Now we can solve the integrals:

\[ \frac{8}{9} \left( u - \ln |u| \right) + C = \frac{8}{9} \left( 1 + 3\sqrt{x} - \ln |1 + 3\sqrt{x}| \right) + C. \]

### Step 6: Substitute back for the second integral

For the second integral, we can use the substitution \(v = \sqrt{x}\), so \(x = v^2\) and \(dx = 2v \, dv\):

\[ \int \frac{6}{v(1 + 3v)} \cdot 2v \, dv = 12 \int \frac{1}{1 + 3v} \, dv = 4 \ln |1 + 3v| + C. \]

### Final Answer

Combining both parts, we have:

\[ \frac{8}{9} \left( 1 + 3\sqrt{x} - \ln |1 + 3\sqrt{x}| \right) + 4 \ln |1 + 3\sqrt{x}| + C. \]

Thus, the final answer is:

\[ \frac{8}{9} + \frac{8}{3}\sqrt{x} + C. \]