`int(3x^(13)+2x^(11))/((2x^4+3x^2+1)^4)dx`

Options

A. `x^4/((2x^4+3x^2+1)^3)+C`
B. `x^12/(6(2x^4+3x^2+1)^3)+C`
C. `x^4/(6(2x^4+3x^2+1)^3)+C`
D. `x^12/((2x^4+3x^2+1)^3)+C`

Answer

To solve the integral

\[ I = \int \frac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Integrand
First, we can simplify the integrand by factoring out \(x^{16}\) from the numerator:

\[ I = \int \frac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4} \, dx = \int \frac{x^{11}(3x^2 + 2)}{(2x^4 + 3x^2 + 1)^4} \, dx. \]

### Step 2: Make a Substitution
Now, let's make a substitution. Let

\[ t = 2 + \frac{3}{x^2} + \frac{1}{x^4}. \]

Next, we need to find \(dt\):

\[ \frac{dt}{dx} = -\frac{6}{x^3} - \frac{4}{x^5} = -\frac{6x^2 + 4}{x^5}. \]

Thus,

\[ dt = -\frac{(6x^2 + 4)}{x^5} \, dx \quad \Rightarrow \quad dx = -\frac{x^5}{6x^2 + 4} \, dt. \]

### Step 3: Substitute in the Integral
Now we substitute \(t\) and \(dx\) into the integral:

\[ I = \int \frac{3x^{13} + 2x^{11}}{t^4} \left(-\frac{x^5}{6x^2 + 4}\right) dt. \]

### Step 4: Simplify the Integral
Now we can express \(3x^{13} + 2x^{11}\) in terms of \(t\):

\[ 3x^{13} + 2x^{11} = x^{11}(3x^2 + 2). \]

Thus, the integral becomes:

\[ I = -\int \frac{x^{11}(3x^2 + 2)}{t^4(6x^2 + 4)} \, dt. \]

### Step 5: Solve the Integral
Now we can integrate:

\[ I = -\frac{1}{6} \int \frac{dt}{t^4}. \]

The integral of \(t^{-4}\) is:

\[ \int t^{-4} \, dt = -\frac{1}{3} t^{-3} + C. \]

Thus,

\[ I = -\frac{1}{6} \left(-\frac{1}{3} t^{-3}\right) + C = \frac{1}{18} t^{-3} + C. \]

### Step 6: Substitute Back for \(t\)
Substituting back for \(t\):

\[ I = \frac{1}{18} \left(2 + \frac{3}{x^2} + \frac{1}{x^4}\right)^{-3} + C. \]

### Final Answer
Thus, the final result for the integral is:

\[ I = \frac{1}{18} \left(2 + \frac{3}{x^2} + \frac{1}{x^4}\right)^{-3} + C. \]