`int(1)/(x sqrt(x^(6)-16))dx=`

Options

A. `sec^(-1)((x^(3))/(4))+c`
B. `(1)/(12)sec^(-1)((x^(3))/(4))+c`
C. `(1)/(3)sec^(-1)((x^(3))/(4))+c`
D. `(1)/(3)sin^(-1)((x^(3))/(4))+c`

Answer

To solve the integral \( \int \frac{1}{x \sqrt{x^6 - 16}} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{1}{x \sqrt{x^6 - 16}} \, dx \]
To simplify the expression, we can multiply the numerator and denominator by \( x^2 \):
\[ \int \frac{x^2}{x^3 \sqrt{x^6 - 16}} \, dx \]

### Step 2: Substitution
Next, we will use the substitution:
\[ t = x^3 \quad \Rightarrow \quad dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
Substituting \( x^2 \, dx \) gives:
\[ x^2 \, dx = \frac{dt}{3} \]
Now, we can rewrite the integral:
\[ \int \frac{1}{x^3 \sqrt{x^6 - 16}} \cdot x^2 \, dx = \int \frac{1}{t \sqrt{t^2 - 16}} \cdot \frac{dt}{3} \]
This simplifies to:
\[ \frac{1}{3} \int \frac{1}{t \sqrt{t^2 - 16}} \, dt \]

### Step 3: Integral of the New Expression
The integral \( \int \frac{1}{t \sqrt{t^2 - 16}} \, dt \) can be solved using the formula:
\[ \int \frac{1}{x \sqrt{x^2 - a^2}} \, dx = \sec^{-1} \left( \frac{x}{a} \right) + C \]
In our case, \( a = 4 \):
\[ \int \frac{1}{t \sqrt{t^2 - 16}} \, dt = \sec^{-1} \left( \frac{t}{4} \right) + C \]

### Step 4: Substitute Back
Now substituting back into our integral:
\[ \frac{1}{3} \left( \sec^{-1} \left( \frac{t}{4} \right) + C \right) = \frac{1}{3} \sec^{-1} \left( \frac{x^3}{4} \right) + C \]

### Step 5: Final Result
Thus, the final result of the integral is:
\[ \int \frac{1}{x \sqrt{x^6 - 16}} \, dx = \frac{1}{3} \sec^{-1} \left( \frac{x^3}{4} \right) + C \]