Evaluate: `int 1/((x+1)(x+2)) dx`

Answer

To evaluate the integral

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

we will use the method of partial fractions. Here’s a step-by-step solution:

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

We want to express the integrand \(\frac{1}{(x+1)(x+2)}\) in the form:

\[ \frac{A}{x+1} + \frac{B}{x+2}, \]

where \(A\) and \(B\) are constants to be determined.

### Step 2: Write the equation

We can write:

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

Multiplying both sides by \((x+1)(x+2)\) to eliminate the denominators gives:

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

### Step 3: Expand and collect terms

Expanding the right-hand side:

\[ 1 = Ax + 2A + Bx + B = (A + B)x + (2A + B). \]

### Step 4: Set up equations for coefficients

For the equation to hold for all \(x\), the coefficients of \(x\) and the constant terms must be equal. This gives us the system of equations:

1. \(A + B = 0\) (coefficient of \(x\))
2. \(2A + B = 1\) (constant term)

### Step 5: Solve the system of equations

From the first equation, we can express \(B\) in terms of \(A\):

\[ B = -A. \]

Substituting \(B = -A\) into the second equation:

\[ 2A - A = 1 \implies A = 1. \]

Now substituting \(A = 1\) back into \(B = -A\):

\[ B = -1. \]

### Step 6: Rewrite the integral

Now we can rewrite the integrand using the values of \(A\) and \(B\):

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

### Step 7: Integrate term by term

Now we can integrate:

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

The integrals can be computed as follows:

\[ \int \frac{1}{x+1} \, dx = \ln|x+1| + C_1, \]
\[ \int \frac{1}{x+2} \, dx = \ln|x+2| + C_2. \]

Thus, we have:

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

### Step 8: Combine the logarithms

Using the properties of logarithms, we can combine the terms:

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

### Final Result

Thus, the final result is:

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