`int(logsqrtx)/(3x)dx` is equal to

Options

A. `(1)/(3)(logsqrtx)^(2)+c`
B. `(2)/(3)(logsqrtx)^(2)+c`
C. `(2)/(3)(logx)^(2)+c`
D. `(1)/(3)(logx)^(2)+c`

Answer

To solve the integral \(\int \frac{\log(\sqrt{x})}{3x} \, dx\), we will follow these steps:

### Step 1: Substitute \( x \) with \( z^2 \)

Let \( x = z^2 \). Then, we differentiate both sides:
\[ dx = 2z \, dz \]

### Step 2: Rewrite the integral

Now we substitute \( x \) and \( dx \) in the integral:
\[ \int \frac{\log(\sqrt{z^2})}{3z^2} \cdot 2z \, dz \]
Since \(\sqrt{z^2} = z\), we can simplify this to:
\[ \int \frac{\log(z)}{3z^2} \cdot 2z \, dz = \int \frac{2 \log(z)}{3z} \, dz \]

### Step 3: Factor out constants

We can factor out the constant \(\frac{2}{3}\):
\[ \frac{2}{3} \int \frac{\log(z)}{z} \, dz \]

### Step 4: Use integration by parts

Let \( u = \log(z) \) and \( dv = \frac{1}{z} \, dz \). Then, we differentiate and integrate:
\[ du = \frac{1}{z} \, dz \quad \text{and} \quad v = \log(z) \]
Using integration by parts, we have:
\[ \int u \, dv = uv - \int v \, du \]
Thus,
\[ \int \log(z) \cdot \frac{1}{z} \, dz = \log(z) \cdot \log(z) - \int \log(z) \cdot \frac{1}{z} \, dz \]
This simplifies to:
\[ \frac{1}{2} (\log(z))^2 \]

### Step 5: Substitute back

Now we substitute back into our integral:
\[ \frac{2}{3} \cdot \frac{1}{2} (\log(z))^2 + C = \frac{1}{3} (\log(z))^2 + C \]

### Step 6: Substitute \( z \) back to \( x \)

Recall that \( z = \sqrt{x} \):
\[ \frac{1}{3} (\log(\sqrt{x}))^2 + C \]

### Final Answer

Thus, the final answer is:
\[ \frac{1}{3} \log^2(\sqrt{x}) + C \]