`int x/(x+a)dx`

Answer

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

### Step 1: Rewrite the integrand
We can rewrite the integrand \( \frac{x}{x+a} \) as follows:
\[ \frac{x}{x+a} = \frac{(x+a) - a}{x+a} = \frac{x+a}{x+a} - \frac{a}{x+a} = 1 - \frac{a}{x+a} \]
Thus, we can express the integral as:
\[ \int \frac{x}{x+a} \, dx = \int \left( 1 - \frac{a}{x+a} \right) \, dx \]

### Step 2: Separate the integral
Now, we can separate the integral into two parts:
\[ \int \left( 1 - \frac{a}{x+a} \right) \, dx = \int 1 \, dx - \int \frac{a}{x+a} \, dx \]

### Step 3: Integrate the first part
The integral of 1 with respect to \( x \) is:
\[ \int 1 \, dx = x \]

### Step 4: Integrate the second part
For the second integral \( \int \frac{a}{x+a} \, dx \), we can factor out the constant \( a \):
\[ \int \frac{a}{x+a} \, dx = a \int \frac{1}{x+a} \, dx \]
The integral \( \int \frac{1}{x+a} \, dx \) is:
\[ \int \frac{1}{x+a} \, dx = \ln |x+a| \]
Thus, we have:
\[ \int \frac{a}{x+a} \, dx = a \ln |x+a| \]

### Step 5: Combine the results
Now, we can combine the results of the two integrals:
\[ \int \frac{x}{x+a} \, dx = x - a \ln |x+a| + C \]
where \( C \) is the constant of integration.

### Final Answer
The final answer to the integral is:
\[ \int \frac{x}{x+a} \, dx = x - a \ln |x+a| + C \]

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