Evaluate :`int(dx)/(xsqrt(x^(6)+1))` equals

Options

A. `sec^(-1)x^(3)+C`
B. `(1)/(6)log((sqrt(x^(6)+1)-1)/(sqrt(x^(6)+1)+1))+C`
C. `(1)/(3)log((sqrt(x^(3)+1)-1)/(sqrt(x^(3)+1)+1))+C`
D. `(1)/(3)log((sqrt(x^(3)+1)+1)/(sqrt(x^(3)+1)-1))+C`

Explanation

To evaluate the integral

\[ I = \int \frac{dx}{x \sqrt{x^6 + 1}}, \]

we can follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integral in a more manageable form. We can multiply and divide the integrand by \(x^5\):

\[ I = \int \frac{x^5}{x^6 \sqrt{x^6 + 1}} \, dx = \int \frac{x^5}{x^6 \sqrt{x^6 + 1}} \, dx. \]

### Step 2: Substitution
Next, we make the substitution \(t^2 = x^6 + 1\). Then, differentiating both sides gives:

\[ 2t \, dt = 6x^5 \, dx \quad \Rightarrow \quad dx = \frac{t \, dt}{3x^5}. \]

Now, we also need to express \(x^5\) in terms of \(t\). From our substitution, we have:

\[ x^6 = t^2 - 1 \quad \Rightarrow \quad x^5 = (t^2 - 1)^{5/6}. \]

### Step 3: Substitute Back into the Integral
Substituting \(dx\) and \(x^5\) back into the integral gives:

\[ I = \int \frac{(t^2 - 1)^{5/6}}{(t^2 - 1)^{3/2} \cdot t} \cdot \frac{t \, dt}{3(t^2 - 1)^{5/6}}. \]

The \( (t^2 - 1)^{5/6} \) terms cancel out, and we are left with:

\[ I = \frac{1}{3} \int \frac{dt}{t^2 - 1}. \]

### Step 4: Integrate
The integral \(\int \frac{dt}{t^2 - 1}\) can be solved using partial fractions:

\[ \frac{1}{t^2 - 1} = \frac{1}{2} \left( \frac{1}{t - 1} - \frac{1}{t + 1} \right). \]

Thus, we have:

\[ I = \frac{1}{3} \cdot \frac{1}{2} \left( \ln |t - 1| - \ln |t + 1| \right) + C = \frac{1}{6} \ln \left| \frac{t - 1}{t + 1} \right| + C. \]

### Step 5: Substitute Back for \(t\)
Now, substituting back \(t = \sqrt{x^6 + 1}\):

\[ I = \frac{1}{6} \ln \left| \frac{\sqrt{x^6 + 1} - 1}{\sqrt{x^6 + 1} + 1} \right| + C. \]

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

\[ I = \frac{1}{6} \ln \left| \frac{\sqrt{x^6 + 1} - 1}{\sqrt{x^6 + 1} + 1} \right| + C. \]