Evaluate the following integrals:
`int (x)/(sqrt(x+1))dx`

Answer

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

### Step 1: Rewrite the integrand
We can rewrite the integrand \( \frac{x}{\sqrt{x+1}} \) by expressing \( x \) as \( (x + 1) - 1 \):
\[ \int \frac{x}{\sqrt{x+1}} \, dx = \int \frac{(x + 1) - 1}{\sqrt{x+1}} \, dx = \int \frac{x + 1}{\sqrt{x+1}} \, dx - \int \frac{1}{\sqrt{x+1}} \, dx \]

### Step 2: Split the integral
Now we can split the integral into two parts:
\[ \int \frac{x}{\sqrt{x+1}} \, dx = \int \frac{x + 1}{\sqrt{x+1}} \, dx - \int \frac{1}{\sqrt{x+1}} \, dx \]
This simplifies to:
\[ \int \sqrt{x+1} \, dx - \int \frac{1}{\sqrt{x+1}} \, dx \]

### Step 3: Evaluate the first integral
For the first integral \( \int \sqrt{x+1} \, dx \), we can use the power rule:
\[ \int \sqrt{x+1} \, dx = \int (x+1)^{1/2} \, dx = \frac{(x+1)^{3/2}}{3/2} = \frac{2}{3} (x+1)^{3/2} \]

### Step 4: Evaluate the second integral
For the second integral \( \int \frac{1}{\sqrt{x+1}} \, dx \), we can also use the power rule:
\[ \int \frac{1}{\sqrt{x+1}} \, dx = \int (x+1)^{-1/2} \, dx = \frac{(x+1)^{1/2}}{-1/2} = -2 (x+1)^{1/2} \]

### Step 5: Combine the results
Now, we can combine the results of both integrals:
\[ \int \frac{x}{\sqrt{x+1}} \, dx = \frac{2}{3} (x+1)^{3/2} - (-2 (x+1)^{1/2}) + C \]
This simplifies to:
\[ \int \frac{x}{\sqrt{x+1}} \, dx = \frac{2}{3} (x+1)^{3/2} + 2 (x+1)^{1/2} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{x}{\sqrt{x+1}} \, dx = \frac{2}{3} (x+1)^{3/2} + 2 (x+1)^{1/2} + C \]