The primitive of `1/(4sqrt(x)+x)` w.r.t.x is

Options

A. `1/2log(sqrt(x)+4)+c`
B. `2log(sqrt(x)+4)+c`
C. `log(sqrt(x)+4)+c`
D. `1/4log(sqrt(x)+4)+c`

Answer

To find the primitive (or integral) of the function \( \frac{1}{4\sqrt{x} + x} \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{4\sqrt{x} + x} \, dx \]

### Step 2: Factor out \( \sqrt{x} \)
Notice that we can factor out \( \sqrt{x} \) from the denominator:
\[ 4\sqrt{x} + x = \sqrt{x}(4 + \sqrt{x}) \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{1}{\sqrt{x}(4 + \sqrt{x})} \, dx \]

### Step 3: Substitute \( u = \sqrt{x} \)
Let \( u = \sqrt{x} \). Then, \( x = u^2 \) and \( dx = 2u \, du \). Substituting these into the integral gives:
\[ I = \int \frac{1}{u(4 + u)} \cdot 2u \, du = 2 \int \frac{1}{4 + u} \, du \]

### Step 4: Integrate
Now we can integrate:
\[ I = 2 \int \frac{1}{4 + u} \, du = 2 \cdot \ln |4 + u| + C \]
where \( C \) is the constant of integration.

### Step 5: Substitute Back
Now we substitute back \( u = \sqrt{x} \):
\[ I = 2 \ln |4 + \sqrt{x}| + C \]

### Final Answer
Thus, the primitive of \( \frac{1}{4\sqrt{x} + x} \) with respect to \( x \) is:
\[ I = 2 \ln |4 + \sqrt{x}| + C \]

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