`int_(1//3)^(3)(1)/(x)log_(e)(|(x+x^(2)-1)/(x-x^(2)+1)|)dx` is equal to

Options

A. `(8)/(3)`
B. `-(8)/(3)`
C. 0
D. 3

Explanation

To solve the integral

\[ I = \int_{\frac{1}{3}}^{3} \frac{1}{x} \log\left|\frac{x + x^2 - 1}{x - x^2 + 1}\right| \, dx, \]

we can use the substitution \( x = \frac{1}{t} \). This substitution will help us simplify the integral.

### Step 1: Substitute \( x = \frac{1}{t} \)

When \( x = \frac{1}{t} \), we have:

\[ dx = -\frac{1}{t^2} dt. \]

The limits of integration will change as follows:

- When \( x = \frac{1}{3} \), \( t = 3 \).
- When \( x = 3 \), \( t = \frac{1}{3} \).

Thus, the integral becomes:

\[ I = \int_{3}^{\frac{1}{3}} \frac{1}{\frac{1}{t}} \log\left|\frac{\frac{1}{t} + \left(\frac{1}{t}\right)^2 - 1}{\frac{1}{t} - \left(\frac{1}{t}\right)^2 + 1}\right| \left(-\frac{1}{t^2}\right) dt. \]

### Step 2: Simplify the integral

This simplifies to:

\[ I = \int_{3}^{\frac{1}{3}} -t \log\left|\frac{\frac{1}{t} + \frac{1}{t^2} - 1}{\frac{1}{t} - \frac{1}{t^2} + 1}\right| \frac{1}{t^2} dt. \]

Rearranging gives:

\[ I = \int_{3}^{\frac{1}{3}} -\frac{1}{t} \log\left|\frac{1 + t - t^2}{1 - t + t^2}\right| dt. \]

### Step 3: Change the limits of integration

Now, we can change the limits of integration:

\[ I = \int_{\frac{1}{3}}^{3} \frac{1}{t} \log\left|\frac{1 + t - t^2}{1 - t + t^2}\right| dt. \]

### Step 4: Combine the integrals

Now we have two expressions for \( I \):

1. \( I = \int_{\frac{1}{3}}^{3} \frac{1}{x} \log\left|\frac{x + x^2 - 1}{x - x^2 + 1}\right| dx \)
2. \( I = \int_{\frac{1}{3}}^{3} \frac{1}{t} \log\left|\frac{1 + t - t^2}{1 - t + t^2}\right| dt \)

### Step 5: Add the two integrals

Adding these two integrals gives:

\[ 2I = \int_{\frac{1}{3}}^{3} \left( \frac{1}{x} \log\left|\frac{x + x^2 - 1}{x - x^2 + 1}\right| + \frac{1}{x} \log\left|\frac{1 + x - x^2}{1 - x + x^2}\right| \right) dx. \]

### Step 6: Simplify the logarithmic expression

Using the property of logarithms, we can combine the logarithmic terms:

\[ 2I = \int_{\frac{1}{3}}^{3} \frac{1}{x} \log\left| \frac{\left(x + x^2 - 1\right)\left(1 + x - x^2\right)}{\left(x - x^2 + 1\right)\left(1 - x + x^2\right)} \right| dx. \]

### Step 7: Evaluate the integral

Notice that the numerator and denominator are symmetric and will cancel out when evaluated over the symmetric limits. Therefore:

\[ 2I = 0 \implies I = 0. \]

### Conclusion

Thus, the value of the integral is

\[ \boxed{0}. \]