`int((x+1)(x+logx)^(2))/(x)dx=?`

Options

A. `(1)/(3)(x+logx)^(3)+C`
B. `(x^(2))/(2)+x+C`
C. `(x^(3))/(3)+(x^(2))/(2)+x+C`
D. none of these

Explanation

To solve the integral

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

we can follow these steps:

### Step 1: Simplify the integrand
First, we can rewrite the integrand by dividing \(x + 1\) by \(x\):

\[ \frac{(x+1)(x+\log x)^{2}}{x} = \left(1 + \frac{1}{x}\right)(x+\log x)^{2}. \]

Now, we can express the integral as:

\[ \int \left(1 + \frac{1}{x}\right)(x+\log x)^{2} \, dx. \]

### Step 2: Distribute the terms
Next, we can distribute the terms inside the integral:

\[ \int (x+\log x)^{2} \, dx + \int \frac{(x+\log x)^{2}}{x} \, dx. \]

### Step 3: Substitution
Now, we will use the substitution \(t = x + \log x\). To differentiate this, we find:

\[ dt = \left(1 + \frac{1}{x}\right) dx. \]

This means that:

\[ dx = \frac{dt}{1 + \frac{1}{x}} = (1 + \frac{1}{x})^{-1} dt. \]

### Step 4: Rewrite the integral
Now we can rewrite the integral in terms of \(t\):

\[ \int (x+\log x)^{2} \, dx = \int t^{2} \, dt. \]

### Step 5: Integrate
Now we can integrate \(t^{2}\):

\[ \int t^{2} \, dt = \frac{t^{3}}{3} + C. \]

### Step 6: Substitute back
Now we substitute back \(t = x + \log x\):

\[ \frac{(x + \log x)^{3}}{3} + C. \]

### Final Answer
Thus, the final answer for the integral is:

\[ \int \frac{(x+1)(x+\log x)^{2}}{x} \, dx = \frac{(x + \log x)^{3}}{3} + C. \]