The value of `int(dx)/(xsqrt(x^(4)-1))` is

Options

A. `1/2sec^(-1)x^(2)+c`
B. `logxsqrt(x^(4)-1)+c`
C. `xlogsqrt(x^(4)-1)+c`
D. `logsqrt(x^(4)-1)+c`

Answer

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

### Step 1: Substitution
Let \( t = x^2 \). Then, differentiating both sides, we get:
\[ dt = 2x \, dx \quad \Rightarrow \quad dx = \frac{dt}{2x} \]
Since \( x = \sqrt{t} \), we can substitute \( dx \) in terms of \( t \):
\[ dx = \frac{dt}{2\sqrt{t}} \]

### Step 2: Rewrite the Integral
Substituting \( dx \) into the integral:
\[ \int \frac{dx}{x \sqrt{x^4 - 1}} = \int \frac{\frac{dt}{2\sqrt{t}}}{\sqrt{t} \sqrt{(\sqrt{t})^4 - 1}} = \int \frac{dt}{2t \sqrt{t^2 - 1}} \]

### Step 3: Simplifying the Integral
Now we have:
\[ \int \frac{dt}{2t \sqrt{t^2 - 1}} \]
This can be simplified to:
\[ \frac{1}{2} \int \frac{dt}{t \sqrt{t^2 - 1}} \]

### Step 4: Recognizing the Integral Form
The integral \( \int \frac{dt}{t \sqrt{t^2 - 1}} \) is a standard integral that evaluates to:
\[ \sec^{-1}(t) + C \]
Thus, we have:
\[ \frac{1}{2} \sec^{-1}(t) + C \]

### Step 5: Back Substituting
Recall that \( t = x^2 \), so we substitute back:
\[ \frac{1}{2} \sec^{-1}(x^2) + C \]

### Final Answer
The value of the integral is:
\[ \int \frac{dx}{x \sqrt{x^4 - 1}} = \frac{1}{2} \sec^{-1}(x^2) + C \]