Evaluate `int (8dx)/((x+2)(x^(2)+4))`

Answer

To evaluate the integral

\[ \int \frac{8 \, dx}{(x+2)(x^2+4)}, \]

we will use the method of partial fractions.

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

We can express the integrand as:

\[ \frac{8}{(x+2)(x^2+4)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+4}. \]

### Step 2: Clear the denominators

Multiplying through by the common denominator \((x+2)(x^2+4)\), we get:

\[ 8 = A(x^2+4) + (Bx+C)(x+2). \]

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

Expanding the right-hand side:

\[ 8 = A(x^2 + 4) + Bx^2 + 2Bx + Cx + 2C. \]

Combining like terms gives:

\[ 8 = (A + B)x^2 + (2B + C)x + (4A + 2C). \]

### Step 4: Set up the system of equations

Now, we can equate the coefficients from both sides:

1. \(A + B = 0\) (coefficient of \(x^2\))
2. \(2B + C = 0\) (coefficient of \(x\))
3. \(4A + 2C = 8\) (constant term)

### Step 5: Solve the system of equations

From equation 1, we have:

\[ B = -A. \]

Substituting \(B = -A\) into equation 2:

\[ 2(-A) + C = 0 \implies C = 2A. \]

Now substituting \(B\) and \(C\) into equation 3:

\[ 4A + 2(2A) = 8 \implies 4A + 4A = 8 \implies 8A = 8 \implies A = 1. \]

Now substituting \(A = 1\) back to find \(B\) and \(C\):

\[ B = -1, \quad C = 2. \]

### Step 6: Write the partial fractions

Thus, we can rewrite the integral as:

\[ \int \left( \frac{1}{x+2} + \frac{-x+2}{x^2+4} \right) dx. \]

### Step 7: Separate the integral

This can be separated into two integrals:

\[ \int \frac{1}{x+2} \, dx - \int \frac{x}{x^2+4} \, dx + 2\int \frac{1}{x^2+4} \, dx. \]

### Step 8: Evaluate each integral

1. The first integral:

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

2. The second integral can be solved using substitution. Let \(u = x^2 + 4\), then \(du = 2x \, dx\) or \(\frac{du}{2} = x \, dx\):

\[ -\int \frac{x}{x^2+4} \, dx = -\frac{1}{2} \int \frac{1}{u} \, du = -\frac{1}{2} \ln|u| = -\frac{1}{2} \ln|x^2 + 4|. \]

3. The third integral:

\[ 2\int \frac{1}{x^2+4} \, dx = 2 \cdot \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) = \tan^{-1}\left(\frac{x}{2}\right). \]

### Step 9: Combine the results

Combining all parts, we have:

\[ \int \frac{8 \, dx}{(x+2)(x^2+4)} = \ln|x+2| - \frac{1}{2} \ln|x^2 + 4| + \tan^{-1}\left(\frac{x}{2}\right) + C. \]

### Final Answer

Thus, the final answer is:

\[ \ln|x+2| - \frac{1}{2} \ln|x^2 + 4| + \tan^{-1}\left(\frac{x}{2}\right) + C. \]