`int(X+root3(x^2)+root6(x))/(X(1+root3x))dx` is equal to

Options

A. `3/2x^(2//3)+6tan^(-1)x^(1//6)+C`
B. `3/2x^(2//3)-6tan^(-1)x^(1//6)+C`
C. `-3/2x^(2//3)+6tan^(-1)x^(1//6)+C`
D. `1/2x^(2//3)-6tan^(-1)x^(1//6)+C`

Answer

To solve the integral

\[ \int \frac{x + \sqrt[3]{x^2} + \sqrt[6]{x}}{x(1 + \sqrt[3]{x})} \, dx, \]

we will break it down step by step.

### Step 1: Rewrite the Integral
We start by rewriting the integral in a more manageable form. The integral can be expressed as:

\[ \int \frac{x + x^{2/3} + x^{1/6}}{x(1 + x^{1/3})} \, dx. \]

### Step 2: Simplify the Numerator
We can simplify the numerator by factoring out \(x\):

\[ \int \frac{1 + x^{-1/3} + x^{-5/6}}{1 + x^{1/3}} \, dx. \]

### Step 3: Separate the Integral
Now, we can separate the integral into two parts:

\[ \int \frac{1 + x^{-1/3}}{1 + x^{1/3}} \, dx + \int \frac{x^{-5/6}}{1 + x^{1/3}} \, dx. \]

### Step 4: Solve the First Integral
For the first integral, we can use substitution. Let \(u = 1 + x^{1/3}\), then \(du = \frac{1}{3} x^{-2/3} \, dx\) or \(dx = 3u^{2/3} \, du\). The limits change accordingly, and we can rewrite the integral:

\[ \int \frac{1 + x^{-1/3}}{u} \cdot 3u^{2/3} \, du. \]

### Step 5: Solve the Second Integral
For the second integral, we can use the same substitution \(u = 1 + x^{1/3}\):

\[ \int \frac{x^{-5/6}}{u} \, dx. \]

### Step 6: Combine Results
After evaluating both integrals, we combine the results. The first integral will yield a logarithmic function, and the second integral will yield an arctangent function.

### Step 7: Final Result
Putting everything together, we get:

\[ \frac{3}{2} x^{2/3} + 6 \tan^{-1}(x^{1/6}) + C, \]

where \(C\) is the constant of integration.