Let `n in N` such that `n gt 1`.
Statement-1: `int_(0)^(oo) (1)/(1+x^(n))dx=int_(0)^(1) (1)/((1-x^(n))^(1//n))dx`
Statement-2: `int_a^b f(x)dx=int_(a)^(b) f(a+b-x)dx`

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 and verify their correctness step by step.

### Step 1: Analyze Statement 1
We need to evaluate the integral \( I_1 = \int_{0}^{\infty} \frac{1}{1+x^n} \, dx \).

**Substitution:**
Let \( x = \tan^{2}(\theta) \). Then, \( dx = 2 \tan(\theta) \sec^{2}(\theta) \, d\theta \).

**Limits of Integration:**
- When \( x = 0 \), \( \theta = 0 \).
- When \( x \to \infty \), \( \theta \to \frac{\pi}{2} \).

**Transforming the Integral:**
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan^{2}(\theta)} \cdot 2 \tan(\theta) \sec^{2}(\theta) \, d\theta \]
Using the identity \( 1 + \tan^{2}(\theta) = \sec^{2}(\theta) \):
\[ I_1 = \int_{0}^{\frac{\pi}{2}} \frac{2 \tan(\theta) \sec^{2}(\theta)}{\sec^{2}(\theta)} \, d\theta = 2 \int_{0}^{\frac{\pi}{2}} \tan(\theta) \, d\theta \]

### Step 2: Evaluate \( \int_{0}^{\frac{\pi}{2}} \tan(\theta) \, d\theta \)
This integral can be evaluated as follows:
\[ \int \tan(\theta) \, d\theta = -\ln|\cos(\theta)| + C \]
Thus,
\[ \int_{0}^{\frac{\pi}{2}} \tan(\theta) \, d\theta = \left[-\ln|\cos(\theta)|\right]_{0}^{\frac{\pi}{2}} = \infty \]
This means \( I_1 \) diverges.

### Step 3: Analyze Statement 2
Statement 2 states that:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \]
This is a well-known property of definite integrals, and it holds true for any continuous function \( f(x) \).

### Step 4: Conclusion
- **Statement 1** is incorrect because the integral diverges.
- **Statement 2** is correct as it is a fundamental property of definite integrals.

Thus, the final conclusion is:
- Statement 1: False
- Statement 2: True