`int(1+x)/(1+3sqrt(x))dx` is equal to

Options

A. `(3)/(5)x^(5//3)+x-(3)/(4)x^(4//3)+x+C`
B. `(3)/(5)x^(5//3)-(3)/(4)x^(4//3)+C`
C. `(3)/(5)x^(5//3)-(3)/(4)x^(4//3)+C`
D. none of these

Answer

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

### Step 1: Rewrite the numerator
We can express \( 1 + x \) in a different form. Notice that:
\[ 1 + x = 1^3 + (3\sqrt{x})^3 \]
This can be rewritten using the identity for the sum of cubes:
\[ a^3 + b^3 = (a+b)(a^2 - ab + b^2) \]
where \( a = 1 \) and \( b = 3\sqrt{x} \).

### Step 2: Apply the sum of cubes formula
Using the sum of cubes formula:
\[ 1 + x = (1 + 3\sqrt{x})\left(1^2 - 1 \cdot 3\sqrt{x} + (3\sqrt{x})^2\right) \]
This simplifies to:
\[ 1 + x = (1 + 3\sqrt{x})(1 - 3\sqrt{x} + 9x) \]

### Step 3: Substitute back into the integral
Now we substitute this back into the integral:
\[ I = \int \frac{(1 + 3\sqrt{x})(1 - 3\sqrt{x} + 9x)}{1 + 3\sqrt{x}} \, dx \]
The \( 1 + 3\sqrt{x} \) terms cancel out:
\[ I = \int (1 - 3\sqrt{x} + 9x) \, dx \]

### Step 4: Integrate each term separately
Now we can integrate each term:
\[ I = \int 1 \, dx - 3\int \sqrt{x} \, dx + 9\int x \, dx \]
Calculating each integral:
1. \( \int 1 \, dx = x \)
2. \( \int \sqrt{x} \, dx = \frac{2}{3}x^{3/2} \)
3. \( \int x \, dx = \frac{1}{2}x^2 \)

Substituting these results back gives:
\[ I = x - 3 \cdot \frac{2}{3}x^{3/2} + 9 \cdot \frac{1}{2}x^2 \]
This simplifies to:
\[ I = x - 2x^{3/2} + \frac{9}{2}x^2 + C \]

### Final Result
Thus, the final result for the integral is:
\[ I = x - 2x^{3/2} + \frac{9}{2}x^2 + C \]