Evaluate : `int(x^(2)-4)/(x^(4)+9x^(2)+16)dx`

Explanation

To evaluate the integral

\[ \int \frac{x^2 - 4}{x^4 + 9x^2 + 16} \, dx, \]

we can follow these steps:

### Step 1: Factor the Denominator
First, we rewrite the denominator \(x^4 + 9x^2 + 16\). We can treat this as a quadratic in terms of \(x^2\):

Let \(u = x^2\), then the expression becomes:

\[ u^2 + 9u + 16. \]

Now we can factor this quadratic. The factors of \(16\) that add up to \(9\) are \(1\) and \(8\):

\[ u^2 + 9u + 16 = (u + 1)(u + 8) = (x^2 + 1)(x^2 + 8). \]

Thus, we can rewrite the integral as:

\[ \int \frac{x^2 - 4}{(x^2 + 1)(x^2 + 8)} \, dx. \]

### Step 2: Partial Fraction Decomposition
Next, we perform partial fraction decomposition on the integrand:

\[ \frac{x^2 - 4}{(x^2 + 1)(x^2 + 8)} = \frac{A}{x^2 + 1} + \frac{B}{x^2 + 8}. \]

Multiplying through by the denominator \((x^2 + 1)(x^2 + 8)\) gives:

\[ x^2 - 4 = A(x^2 + 8) + B(x^2 + 1). \]

Expanding the right-hand side:

\[ x^2 - 4 = Ax^2 + 8A + Bx^2 + B = (A + B)x^2 + (8A + B). \]

### Step 3: Set Up the System of Equations
Now we can set up the system of equations by equating coefficients:

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

### Step 4: Solve the System
From the first equation, we can express \(B\) in terms of \(A\):

\[ B = 1 - A. \]

Substituting into the second equation:

\[ 8A + (1 - A) = -4, \]
\[ 7A + 1 = -4, \]
\[ 7A = -5 \implies A = -\frac{5}{7}. \]

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

\[ B = 1 - \left(-\frac{5}{7}\right) = 1 + \frac{5}{7} = \frac{12}{7}. \]

### Step 5: Rewrite the Integral
Now we can rewrite the integral as:

\[ \int \left( \frac{-\frac{5}{7}}{x^2 + 1} + \frac{\frac{12}{7}}{x^2 + 8} \right) \, dx. \]

This can be separated into two integrals:

\[ -\frac{5}{7} \int \frac{1}{x^2 + 1} \, dx + \frac{12}{7} \int \frac{1}{x^2 + 8} \, dx. \]

### Step 6: Evaluate the Integrals
The integrals can be evaluated as follows:

1. \(\int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) + C\).
2. \(\int \frac{1}{x^2 + 8} \, dx = \frac{1}{\sqrt{8}} \tan^{-1}\left(\frac{x}{\sqrt{8}}\right) + C = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x}{2\sqrt{2}}\right) + C\).

Putting it all together, we have:

\[ -\frac{5}{7} \tan^{-1}(x) + \frac{12}{7} \cdot \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x}{2\sqrt{2}}\right) + C. \]

### Final Answer
Thus, the final answer is:

\[ -\frac{5}{7} \tan^{-1}(x) + \frac{6}{7\sqrt{2}} \tan^{-1}\left(\frac{x}{2\sqrt{2}}\right) + C. \]