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 gr 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 will analyze both statements step by step.

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

**Step 1: Change of Variable**
Let \( t = \frac{1}{u} \). Then, we have:
\[ dt = -\frac{1}{u^2} du. \]
Now, we need to change the limits of integration. When \( t = 1 \), \( u = 1 \) and when \( t = x \), \( u = \frac{1}{x} \).

**Step 2: Substitute in the Integral**
Substituting the variable in the integral, we get:
\[ f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}} \frac{\log_e\left(\frac{1}{u}\right)}{1 + \frac{1}{u} + \left(\frac{1}{u}\right)^2} \left(-\frac{1}{u^2}\right) du. \]
This simplifies to:
\[ f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}} \frac{-\log_e u}{\frac{u^2 + u + 1}{u^2}} \left(-\frac{1}{u^2}\right) du. \]
Further simplifying gives:
\[ f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}} \frac{\log_e u}{1 + u + u^2} du. \]

**Step 3: Change the Limits of Integration**
Now, we can change the limits of integration:
\[ f\left(\frac{1}{x}\right) = \int_{\frac{1}{x}}^{1} \frac{\log_e u}{1 + u + u^2} du. \]
By reversing the limits, we get:
\[ f\left(\frac{1}{x}\right) = -\int_{1}^{\frac{1}{x}} \frac{\log_e u}{1 + u + u^2} du. \]
Thus, we can conclude:
\[ f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_e t}{1 + t + t^2} dt = f(x). \]

### Conclusion for Statement 1:
Thus, we have shown that \( f(x) = f\left(\frac{1}{x}\right) \) for all \( x > 0 \).

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

**Step 1: Change of Variable for \( f\left(\frac{1}{x}\right) \)**
Using the same substitution as before, let \( t = \frac{1}{u} \):
\[ f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}} \frac{\log_e\left(\frac{1}{u}\right)}{1 + \frac{1}{u}} \left(-\frac{1}{u^2}\right) du. \]
This simplifies to:
\[ f\left(\frac{1}{x}\right) = \int_{1}^{\frac{1}{x}} \frac{-\log_e u}{\frac{u + 1}{u}} \left(-\frac{1}{u^2}\right) du = \int_{1}^{\frac{1}{x}} \frac{\log_e u}{u + 1} du. \]

**Step 2: Combine the Two Integrals**
Now we combine \( f(x) \) and \( f\left(\frac{1}{x}\right) \):
\[ f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_e t}{1 + t} dt + \int_{1}^{\frac{1}{x}} \frac{\log_e u}{1 + u} du. \]
Changing the limits of the second integral gives:
\[ f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_e t}{1 + t} dt + \int_{\frac{1}{x}}^{1} \frac{\log_e u}{1 + u} du. \]

**Step 3: Final Calculation**
Combining these integrals leads to:
\[ f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log_e t}{1 + t} dt + \int_{1}^{x} \frac{-\log_e t}{1 + t} dt = \int_{1}^{x} \frac{\log_e t}{1 + t} dt + \int_{1}^{x} \frac{\log_e t}{1 + t} dt = \frac{(\log_e x)^2}{2}. \]

### Conclusion for Statement 2:
Thus, we have shown that \( f(x) + f\left(\frac{1}{x}\right) = \frac{(\log_e x)^2}{2} \).

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### Final Conclusion:
Both statements are true, but Statement 2 does not serve as a correct explanation for Statement 1, as they are independent results.

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