The value of `int(dx)/(sqrt(x)+root3(x))` is

Options

A. `root3(x)+3(root3(x))-6(root6(x))+6log(root6(x)+1)+C`
B. `2sqrt(x)+root3(x)-6log(root6(x+1))+C`
C. `2sqrt(x)-3root3(x)+6root6(x)-6log(root6 (x)+1)+C`
D. none of these

Answer

To solve the integral

\[ I = \int \frac{dx}{\sqrt{x} + \sqrt[3]{x}} \]

we will use a substitution method. Let's break it down step by step.

### Step 1: Substitute Variables
Let \( u = x^{1/6} \). Then, we have:
\[ x = u^6 \quad \text{and} \quad dx = 6u^5 \, du \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral in terms of \( u \):
\[ I = \int \frac{6u^5 \, du}{\sqrt{u^6} + \sqrt[3]{u^6}} = \int \frac{6u^5 \, du}{u^3 + u^2} \]

### Step 3: Simplify the Denominator
Factor out \( u^2 \) from the denominator:
\[ I = \int \frac{6u^5 \, du}{u^2(u + 1)} = 6 \int \frac{u^3 \, du}{u + 1} \]

### Step 4: Polynomial Long Division
Now, perform polynomial long division on \( \frac{u^3}{u + 1} \):
- Divide \( u^3 \) by \( u + 1 \):
- \( u^2 \) gives \( u^3 + u^2 \)
- Subtract: \( u^3 - (u^3 + u^2) = -u^2 \)
- Next, divide \( -u^2 \) by \( u + 1 \):
- \( -u \) gives \( -u^2 - u \)
- Subtract: \( -u^2 - (-u^2 - u) = u \)
- Finally, divide \( u \) by \( u + 1 \):
- \( 1 \) gives \( u + 1 \)
- Subtract: \( u - (u + 1) = -1 \)

Thus, we have:
\[ \frac{u^3}{u + 1} = u^2 - u + 1 - \frac{1}{u + 1} \]

### Step 5: Substitute Back into the Integral
Now substitute this back into the integral:
\[ I = 6 \int \left( u^2 - u + 1 - \frac{1}{u + 1} \right) du \]

### Step 6: Integrate Each Term
Now integrate each term:
\[ I = 6 \left( \frac{u^3}{3} - \frac{u^2}{2} + u - \log|u + 1| \right) + C \]

### Step 7: Substitute Back for \( u \)
Substituting back \( u = x^{1/6} \):
\[ I = 6 \left( \frac{(x^{1/6})^3}{3} - \frac{(x^{1/6})^2}{2} + x^{1/6} - \log|x^{1/6} + 1| \right) + C \]
\[ = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6\log|x^{1/6} + 1| + C \]

### Final Answer
Thus, the value of the integral is:
\[ I = 2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} - 6\log(\sqrt[6]{x} + 1) + C \]