`int1/(x^(6)+x^(4))dx` is equal to

Options

A. `1/(3x^(3))-1/x+tan^(-1)x+c`
B. `1/(3x^(3))+1/x-tan^(-1)x+c`
C. `-1/(3x^(3))-1/x+tan^(-1)x+c`
D. `-1/(3x^(3))+1/x+tan^(-1)x+c`

Answer

To solve the integral \( \int \frac{1}{x^6 + x^4} \, dx \), we can follow these steps:

### Step 1: Factor the Denominator
We start by factoring the denominator:
\[ x^6 + x^4 = x^4(x^2 + 1) \]
Thus, we can rewrite the integral as:
\[ \int \frac{1}{x^4(x^2 + 1)} \, dx \]

### Step 2: Use Partial Fraction Decomposition
Next, we will use partial fraction decomposition to break down the integrand:
\[ \frac{1}{x^4(x^2 + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x^4} + \frac{Ex + F}{x^2 + 1} \]
Multiplying through by the denominator \( x^4(x^2 + 1) \) gives:
\[ 1 = A x^3(x^2 + 1) + B x^2(x^2 + 1) + C x(x^2 + 1) + D(x^2 + 1) + (Ex + F)x^4 \]

### Step 3: Expand and Collect Like Terms
Expanding the right-hand side, we collect like terms based on powers of \( x \):
\[ 1 = (A + B + C + D)x^4 + (E)x^4 + (A + B + C + D)x^2 + (F)x^0 \]
Setting coefficients equal gives us a system of equations.

### Step 4: Solve for Coefficients
By substituting values for \( x \) (like \( x = 0 \)), we can find the coefficients \( A, B, C, D, E, F \). After solving, we find:
- \( D = 1 \)
- \( B = -1 \)
- \( A = 0 \)
- \( C = 0 \)
- \( E = 0 \)
- \( F = 0 \)

Thus, the partial fraction decomposition simplifies to:
\[ \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1} \]

### Step 5: Integrate Each Term
Now we can integrate each term separately:
\[ \int \left( \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1} \right) \, dx = \int x^{-4} \, dx - \int x^{-2} \, dx + \int \frac{1}{x^2 + 1} \, dx \]

Calculating these integrals:
1. \( \int x^{-4} \, dx = -\frac{1}{3x^3} \)
2. \( \int x^{-2} \, dx = -\frac{1}{x} \)
3. \( \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) \)

### Step 6: Combine Results
Combining the results of the integrals, we have:
\[ -\frac{1}{3x^3} + \frac{1}{x} + \tan^{-1}(x) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{1}{x^6 + x^4} \, dx = -\frac{1}{3x^3} + \frac{1}{x} + \tan^{-1}(x) + C \]