`int(x^(9))/((4x^(2)+ 1)^(6))dx` is equal to

Options

A. `(1)/(10) ( 4+(1)/(x^(2)))^(-5)+C`
B. `(1)/(10x) ( 4+(1)/(x^(2)))^(-5)+C`
C. `(1)/(5) ( 4+(1)/(x^(2)))^(-5)+C`
D. `(1)/(5x) ( 4+(1)/(x^(2)))^(-5)+C`

Answer

To solve the integral

\[ \int \frac{x^9}{(4x^2 + 1)^6} \, dx, \]

we can follow these steps:

### Step 1: Simplify the Denominator
We can factor out \(x^2\) from the denominator:

\[ (4x^2 + 1)^6 = \left(x^2(4 + \frac{1}{x^2})\right)^6 = (x^2)^6 (4 + \frac{1}{x^2})^6 = x^{12} (4 + \frac{1}{x^2})^6. \]

### Step 2: Rewrite the Integral
Now we can rewrite the integral as:

\[ \int \frac{x^9}{x^{12} (4 + \frac{1}{x^2})^6} \, dx = \int \frac{1}{x^3 (4 + \frac{1}{x^2})^6} \, dx. \]

### Step 3: Substitution
Let \(t = 4 + \frac{1}{x^2}\). Then, we differentiate \(t\):

\[ dt = -\frac{2}{x^3} \, dx \implies dx = -\frac{x^3}{2} dt. \]

### Step 4: Substitute in the Integral
Now, we substitute \(dx\) and \(t\) into the integral:

\[ \int \frac{1}{x^3 t^6} \left(-\frac{x^3}{2} dt\right) = -\frac{1}{2} \int t^{-6} \, dt. \]

### Step 5: Integrate
Now we can integrate:

\[ -\frac{1}{2} \int t^{-6} \, dt = -\frac{1}{2} \cdot \frac{t^{-5}}{-5} + C = \frac{1}{10} t^{-5} + C. \]

### Step 6: Substitute Back
Substituting back \(t = 4 + \frac{1}{x^2}\):

\[ \frac{1}{10} \left(4 + \frac{1}{x^2}\right)^{-5} + C. \]

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

\[ \int \frac{x^9}{(4x^2 + 1)^6} \, dx = \frac{1}{10} \left(4 + \frac{1}{x^2}\right)^{-5} + C. \]