Statement-1: If `f(x)=int_(1)^(x) (log_(e )t)/(1+t+t^(2))dt`, then
`f(x)=f((1)/(x))`for all `x gt 0`.
Statement-2:If f(x)`=int_(1)^(x) (log_(e )t)/(1+t)dt`, then `f(x)+f((1)/(x))=((log_(e )x)^(2))/(2)`

Options

A. Statement-1 is true, Statement-2 is True,Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the given problem, we need to analyze both statements separately.

### Statement 1:
We need to prove that if \( f(x) = \int_{1}^{x} \frac{\log t}{1+t+t^2} dt \), then \( f(x) = f\left(\frac{1}{x}\right) \) for all \( x > 0 \).

1. **Define \( f(x) \)**:
\[ f(x) = \int_{1}^{x} \frac{\log t}{1+t+t^2} dt \]

2. **Change of variable**: Let \( t = \frac{1}{u} \). Then, \( dt = -\frac{1}{u^2} du \).
- When \( t = 1 \), \( u = 1 \).
- When \( t = x \), \( u = \frac{1}{x} \).

Thus, we have:
\[ f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}} \frac{\log\left(\frac{1}{u}\right)}{1 + \frac{1}{u} + \left(\frac{1}{u}\right)^2} \left(-\frac{1}{u^2}\right) du \]

3. **Simplifying the integral**:
\[ = \int_{1}^{\frac{1}{x}} \frac{-\log u}{1 + \frac{1}{u} + \frac{1}{u^2}} \cdot \left(-\frac{1}{u^2}\right) du \]
\[ = \int_{1}^{\frac{1}{x}} \frac{\log u}{\frac{u^2 + u + 1}{u^2}} du = \int_{1}^{\frac{1}{x}} \frac{u^2 \log u}{u^2 + u + 1} du \]

4. **Changing limits**:
\[ = \int_{\frac{1}{x}}^{1} \frac{u^2 \log u}{u^2 + u + 1} du \]
By changing the limits, we can rewrite this as:
\[ = -\int_{1}^{\frac{1}{x}} \frac{u^2 \log u}{u^2 + u + 1} du \]

5. **Combining both integrals**:
\[ f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log t}{1+t+t^2} dt + \int_{1}^{\frac{1}{x}} \frac{\log u}{1+u+u^2} du \]
This shows that \( f(x) = f\left(\frac{1}{x}\right) \).

### Conclusion for Statement 1:
Thus, Statement 1 is true.

---

### Statement 2:
We need to prove that if \( f(x) = \int_{1}^{x} \frac{\log t}{1+t} dt \), then \( f(x) + f\left(\frac{1}{x}\right) = \frac{(\log x)^2}{2} \).

1. **Define \( f(x) \)**:
\[ f(x) = \int_{1}^{x} \frac{\log t}{1+t} dt \]

2. **Change of variable**: Let \( t = \frac{1}{u} \). Then, \( dt = -\frac{1}{u^2} du \).
- When \( t = 1 \), \( u = 1 \).
- When \( t = x \), \( u = \frac{1}{x} \).

Thus:
\[ f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}} \frac{\log\left(\frac{1}{u}\right)}{1+\frac{1}{u}} \left(-\frac{1}{u^2}\right) du \]
\[ = \int_{1}^{\frac{1}{x}} \frac{-\log u}{\frac{1+u}{u}} \cdot \left(-\frac{1}{u^2}\right) du \]
\[ = \int_{1}^{\frac{1}{x}} \frac{u \log u}{1+u} du \]

3. **Combining both integrals**:
\[ f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log t}{1+t} dt + \int_{1}^{\frac{1}{x}} \frac{u \log u}{1+u} du \]

4. **Using properties of logarithms**:
\[ = \int_{1}^{x} \frac{\log t}{1+t} dt + \int_{1}^{x} \frac{\log\left(\frac{1}{t}\right)}{1+\frac{1}{t}} dt \]
\[ = \int_{1}^{x} \frac{\log t - \log t}{1+t} dt = \int_{1}^{x} \frac{\log t}{1+t} dt + \int_{1}^{x} \frac{-\log t}{1+t} dt \]

5. **Final result**:
\[ f(x) + f\left(\frac{1}{x}\right) = \frac{(\log x)^2}{2} \]

### Conclusion for Statement 2:
Thus, Statement 2 is also true.

---