The value of `int_(1)^(7sqrt(2)) (1)/(x(2x^(7)+1)dx` is

Options

A. `log(6//5)`
B. `6log(6//5)`
C. `(1//7)log(6//5)`
D. `(1//12)log(6//5)`

Answer

To evaluate the integral

\[ I = \int_{1}^{7\sqrt{2}} \frac{1}{x(2x^7 + 1)} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the integrand. We can factor out \(x^7\) from the denominator:

\[ I = \int_{1}^{7\sqrt{2}} \frac{1}{x \left( x^7(2 + \frac{1}{x^7}) \right)} \, dx = \int_{1}^{7\sqrt{2}} \frac{1}{x^8 \left( 2 + \frac{1}{x^7} \right)} \, dx. \]

### Step 2: Substitution
Next, we will use the substitution:

\[ t = 2 + \frac{1}{x^7}. \]

Differentiating both sides gives:

\[ dt = -\frac{7}{x^8} \, dx \quad \Rightarrow \quad dx = -\frac{x^8}{7} \, dt. \]

### Step 3: Change the Limits
Now we need to change the limits of integration.

- When \(x = 1\):
\[ t = 2 + \frac{1}{1^7} = 3. \]
- When \(x = 7\sqrt{2}\):
\[ t = 2 + \frac{1}{(7\sqrt{2})^7} = 2 + \frac{1}{7^7 \cdot 2^{7/2}}. \]

### Step 4: Substitute in the Integral
Substituting \(t\) and \(dx\) into the integral gives:

\[ I = \int_{3}^{2 + \frac{1}{(7\sqrt{2})^7}} \frac{-1}{7t} \, dt. \]

### Step 5: Adjust the Integral
Since we have a negative sign, we can switch the limits:

\[ I = \frac{1}{7} \int_{2 + \frac{1}{(7\sqrt{2})^7}}^{3} \frac{1}{t} \, dt. \]

### Step 6: Evaluate the Integral
The integral of \(\frac{1}{t}\) is \(\ln t\):

\[ I = \frac{1}{7} \left[ \ln t \right]_{2 + \frac{1}{(7\sqrt{2})^7}}^{3} = \frac{1}{7} \left( \ln 3 - \ln \left( 2 + \frac{1}{(7\sqrt{2})^7} \right) \right). \]

### Step 7: Simplify Using Logarithmic Properties
Using the property \(\ln a - \ln b = \ln \frac{a}{b}\):

\[ I = \frac{1}{7} \ln \left( \frac{3}{2 + \frac{1}{(7\sqrt{2})^7}} \right). \]

### Final Answer
Thus, the value of the integral is:

\[ I = \frac{1}{7} \ln \left( \frac{3}{2 + \frac{1}{(7\sqrt{2})^7}} \right). \]