`int (dx)/((x+1)(x+2)(x+3))`

Answer

To solve the integral

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

we will use the method of partial fractions.

### Step 1: Set up the partial fraction decomposition

We can express the integrand as a sum of simpler fractions:

\[ \frac{1}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}. \]

### Step 2: Multiply through by the denominator

To eliminate the denominators, multiply both sides by \((x+1)(x+2)(x+3)\):

\[ 1 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2). \]

### Step 3: Expand the right-hand side

Expanding the right-hand side gives:

\[ 1 = A(x^2 + 5x + 6) + B(x^2 + 4x + 3) + C(x^2 + 3x + 2). \]

### Step 4: Combine like terms

Combining like terms, we have:

\[ 1 = (A + B + C)x^2 + (5A + 4B + 3C)x + (6A + 3B + 2C). \]

### Step 5: Set up a system of equations

Since this must hold for all \(x\), we can equate coefficients:

1. \(A + B + C = 0\) (coefficient of \(x^2\))
2. \(5A + 4B + 3C = 0\) (coefficient of \(x\))
3. \(6A + 3B + 2C = 1\) (constant term)

### Step 6: Solve the system of equations

From equation (1), we can express \(C\) in terms of \(A\) and \(B\):

\[ C = -A - B. \]

Substituting \(C\) into equations (2) and (3):

1. \(5A + 4B + 3(-A - B) = 0\) simplifies to \(2A + B = 0\) or \(B = -2A\).
2. Substituting \(B = -2A\) into equation (3):

\[ 6A + 3(-2A) + 2(-A - (-2A)) = 1 \implies 6A - 6A + 2A = 1 \implies 2A = 1 \implies A = \frac{1}{2}. \]

Now substituting \(A\) back to find \(B\) and \(C\):

\[ B = -2 \cdot \frac{1}{2} = -1, \]
\[ C = -\frac{1}{2} - (-1) = \frac{1}{2}. \]

### Step 7: Write the partial fraction decomposition

Now we have:

\[ \frac{1}{(x+1)(x+2)(x+3)} = \frac{1/2}{x+1} - \frac{1}{x+2} + \frac{1/2}{x+3}. \]

### Step 8: Integrate each term

Now we can integrate each term separately:

\[ \int \left( \frac{1/2}{x+1} - \frac{1}{x+2} + \frac{1/2}{x+3} \right) dx = \frac{1}{2} \ln|x+1| - \ln|x+2| + \frac{1}{2} \ln|x+3| + C. \]

### Step 9: Combine the logarithms

Combining the logarithmic terms, we have:

\[ \frac{1}{2} \ln|x+1| + \frac{1}{2} \ln|x+3| - \ln|x+2| = \ln\left(\sqrt{(x+1)(x+3)}\right) - \ln|x+2| = \ln\left(\frac{\sqrt{(x+1)(x+3)}}{|x+2|}\right). \]

### Final Answer

Thus, the integral evaluates to:

\[ \int \frac{dx}{(x+1)(x+2)(x+3)} = \ln\left(\frac{\sqrt{(x+1)(x+3)}}{|x+2|}\right) + C. \]