`int (1)/(sqrtx+ sqrt(x^(3)))dx=`….

Options

A. `2 tan^(-1) (sqrtx)+c`
B. `tan^(-1) (2x-1)+c`
C. `2 tan sqrtx+c`
D. `2 tan^(-1) (x) +c`

Answer

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

### Step 1: Simplify the integrand
We start by rewriting the integrand:
\[ \sqrt{x^3} = \sqrt{x} \cdot \sqrt{x^2} = \sqrt{x} \cdot x = x^{3/2} \]
Thus, we can express the integral as:
\[ \int \frac{1}{\sqrt{x} + x^{3/2}} \, dx \]
Next, we can factor out \( \sqrt{x} \) from the denominator:
\[ \sqrt{x} + x^{3/2} = \sqrt{x}(1 + \sqrt{x}) \]
So, we rewrite the integral:
\[ \int \frac{1}{\sqrt{x}(1 + \sqrt{x})} \, dx \]

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

### Step 3: Integrate
Now we can integrate:
\[ \int \frac{2}{1 + t} \, dt = 2 \ln |1 + t| + C \]

### Step 4: Substitute back for \( t \)
Recall that \( t = \sqrt{x} \). Therefore, we substitute back:
\[ 2 \ln |1 + \sqrt{x}| + C \]

### Final Answer
The final result of the integral is:
\[ \int \frac{1}{\sqrt{x} + \sqrt{x^3}} \, dx = 2 \ln |1 + \sqrt{x}| + C \]

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