If `f(x)=int_(1)^(x)(logt)(1+t+t^(2))dt AAxge1`, then prove that `f(x)=f(1/x)`.

Explanation

To prove that \( f(x) = f\left(\frac{1}{x}\right) \) for the function defined as

\[ f(x) = \int_{1}^{x} \frac{\log t}{1+t+t^2} \, dt \quad \text{for } x > 1, \]

we will start by calculating \( f\left(\frac{1}{x}\right) \).

### Step 1: Write down \( f\left(\frac{1}{x}\right) \)

We have:

\[ f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}} \frac{\log t}{1+t+t^2} \, dt. \]

### Step 2: Change of variable

Now, we will perform a change of variable. Let \( t = \frac{1}{y} \). Then, we have:

\[ dt = -\frac{1}{y^2} \, dy. \]

### Step 3: Change the limits of integration

When \( t = 1 \), \( y = 1 \). When \( t = \frac{1}{x} \), \( y = x \). Thus, the limits of integration change from \( 1 \) to \( \frac{1}{x} \) to \( 1 \) to \( x \).

### Step 4: Substitute and simplify

Now substituting in the integral, we get:

\[ f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log\left(\frac{1}{y}\right)}{1+\frac{1}{y}+\left(\frac{1}{y}\right)^2} \left(-\frac{1}{y^2}\right) dy. \]

### Step 5: Simplify the logarithm and the denominator

The logarithm simplifies as follows:

\[ \log\left(\frac{1}{y}\right) = -\log y. \]

The denominator simplifies to:

\[ 1 + \frac{1}{y} + \frac{1}{y^2} = \frac{y^2 + y + 1}{y^2}. \]

Thus, we have:

\[ f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{-\log y}{\frac{y^2 + y + 1}{y^2}} \left(-\frac{1}{y^2}\right) dy = \int_{1}^{x} \frac{\log y}{1 + y + y^2} dy. \]

### Step 6: Conclude the proof

This shows that:

\[ f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log y}{1 + y + y^2} \, dy = f(x). \]

Thus, we conclude that:

\[ f(x) = f\left(\frac{1}{x}\right). \]