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

Answer

To evaluate the integral

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

we can follow these steps:

### Step 1: Simplify the Integral
First, we can factor out \(x^2\) from both the numerator and the denominator:

\[ \int \frac{x^2 + 4}{x^4 + 16} \, dx = \int \frac{x^2(1 + \frac{4}{x^2})}{x^2(x^2 + \frac{16}{x^2})} \, dx. \]

This simplifies to:

\[ \int \frac{1 + \frac{4}{x^2}}{x^2 + \frac{16}{x^2}} \, dx. \]

### Step 2: Rewrite the Denominator
Next, we rewrite the denominator to make it easier to integrate. We can express \(x^2 + \frac{16}{x^2}\) as a perfect square:

\[ x^2 + \frac{16}{x^2} = \left(x - \frac{4}{x}\right)^2 + 8. \]

### Step 3: Substitute for Simplification
Let \(t = x - \frac{4}{x}\). Then we differentiate \(t\) with respect to \(x\):

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

Now we can rewrite the integral in terms of \(t\):

\[ \int \frac{dt}{t^2 + 8}. \]

### Step 4: Integrate
The integral \(\int \frac{dt}{t^2 + 8}\) can be solved using the formula for the integral of the form \(\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\):

Here, \(a^2 = 8\) implies \(a = 2\sqrt{2}\):

\[ \int \frac{dt}{t^2 + 8} = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{t}{2\sqrt{2}}\right) + C. \]

### Step 5: Substitute Back
Now, we substitute back \(t = x - \frac{4}{x}\):

\[ \int \frac{x^2 + 4}{x^4 + 16} \, dx = \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x - \frac{4}{x}}{2\sqrt{2}}\right) + C. \]

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

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