`int x^4/(x^4-16) dx`

Answer

To solve the integral \( \int \frac{x^4}{x^4 - 16} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We can rewrite the integrand as follows:
\[ \frac{x^4}{x^4 - 16} = 1 + \frac{16}{x^4 - 16} \]
Thus, we can express the integral as:
\[ \int \frac{x^4}{x^4 - 16} \, dx = \int 1 \, dx + \int \frac{16}{x^4 - 16} \, dx \]

### Step 2: Integrate the first part
The first integral is straightforward:
\[ \int 1 \, dx = x \]

### Step 3: Factor the denominator of the second integral
Now, we need to factor \( x^4 - 16 \):
\[ x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4) \]
So, we can rewrite the second integral as:
\[ \int \frac{16}{(x - 2)(x + 2)(x^2 + 4)} \, dx \]

### Step 4: Use partial fraction decomposition
We will express \( \frac{16}{(x - 2)(x + 2)(x^2 + 4)} \) in terms of partial fractions:
\[ \frac{16}{(x - 2)(x + 2)(x^2 + 4)} = \frac{A}{x - 2} + \frac{B}{x + 2} + \frac{Cx + D}{x^2 + 4} \]
Multiplying through by the denominator gives:
\[ 16 = A(x + 2)(x^2 + 4) + B(x - 2)(x^2 + 4) + (Cx + D)(x - 2)(x + 2) \]

### Step 5: Solve for coefficients
To find \( A, B, C, D \), we can substitute convenient values for \( x \) (like \( x = 2 \) and \( x = -2 \)) and also compare coefficients.

1. Let \( x = 2 \):
\[ 16 = A(4)(8) \Rightarrow A = \frac{1}{8} \]
2. Let \( x = -2 \):
\[ 16 = B(-4)(8) \Rightarrow B = -\frac{1}{8} \]
3. For \( C \) and \( D \), we can compare coefficients of \( x^3, x^2, x^1, x^0 \) from both sides.

After solving, we find:
\[ A = \frac{1}{8}, \quad B = -\frac{1}{8}, \quad C = 0, \quad D = 2 \]

### Step 6: Rewrite the integral
Now we can rewrite the integral:
\[ \int \frac{16}{(x - 2)(x + 2)(x^2 + 4)} \, dx = \int \left( \frac{1/8}{x - 2} - \frac{1/8}{x + 2} + \frac{2}{x^2 + 4} \right) \, dx \]

### Step 7: Integrate each term
Now we integrate each term:
1. \( \int \frac{1/8}{x - 2} \, dx = \frac{1}{8} \ln |x - 2| \)
2. \( \int -\frac{1/8}{x + 2} \, dx = -\frac{1}{8} \ln |x + 2| \)
3. \( \int \frac{2}{x^2 + 4} \, dx = \frac{2}{2} \tan^{-1} \left( \frac{x}{2} \right) = \tan^{-1} \left( \frac{x}{2} \right) \)

### Step 8: Combine results
Combining all parts, we have:
\[ \int \frac{x^4}{x^4 - 16} \, dx = x + \frac{1}{8} \ln |x - 2| - \frac{1}{8} \ln |x + 2| + \tan^{-1} \left( \frac{x}{2} \right) + C \]

### Final Answer:
\[ \int \frac{x^4}{x^4 - 16} \, dx = x + \frac{1}{8} \ln \left| \frac{x - 2}{x + 2} \right| + \tan^{-1} \left( \frac{x}{2} \right) + C \]