Evaluate the following integrals:
`int x sqrt(x+2)dx`

Answer

To evaluate the integral \( \int x \sqrt{x+2} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int x \sqrt{x+2} \, dx \]
We can express \( x \) in terms of \( (x + 2) \):
\[ x = (x + 2) - 2 \]
Thus, we can rewrite the integral as:
\[ \int ((x + 2) - 2) \sqrt{x + 2} \, dx \]
This expands to:
\[ \int (x + 2) \sqrt{x + 2} \, dx - 2 \int \sqrt{x + 2} \, dx \]

### Step 2: Change of Variables
Let \( u = x + 2 \). Then \( du = dx \) and \( x = u - 2 \). The integral becomes:
\[ \int (u) \sqrt{u} \, du - 2 \int \sqrt{u} \, du \]

### Step 3: Simplify the Integrals
Now we can simplify the integrals:
1. For the first integral:
\[ \int u \sqrt{u} \, du = \int u^{3/2} \, du \]
2. For the second integral:
\[ \int \sqrt{u} \, du = \int u^{1/2} \, du \]

### Step 4: Evaluate the Integrals
Now we apply the power rule for integration:
1. For \( \int u^{3/2} \, du \):
\[ = \frac{u^{3/2 + 1}}{3/2 + 1} = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2} \]
2. For \( \int u^{1/2} \, du \):
\[ = \frac{u^{1/2 + 1}}{1/2 + 1} = \frac{u^{3/2}}{3/2} = \frac{2}{3} u^{3/2} \]

### Step 5: Combine the Results
Substituting back into the expression, we have:
\[ \int x \sqrt{x+2} \, dx = \frac{2}{5} u^{5/2} - 2 \cdot \frac{2}{3} u^{3/2} + C \]
Substituting \( u = x + 2 \):
\[ = \frac{2}{5} (x + 2)^{5/2} - \frac{4}{3} (x + 2)^{3/2} + C \]

### Final Answer
Thus, the final result is:
\[ \int x \sqrt{x+2} \, dx = \frac{2}{5} (x + 2)^{5/2} - \frac{4}{3} (x + 2)^{3/2} + C \]