Statement I `int((1)/(1+x^(4)))dx=tan^(-1)(x^(2))+C`
Statement II `int(1)/(1+x^(2))dx=tan^(-1)x +C`

Options

A. Statement I is true, Statement II is also true , Statement II is the correct explanation of Statement I.
B. Statement I is true, Statement II is also true, Statement II is not the correct explanation of Statement I.
C. Statement I is true, Statement II is false.
D. Statement I is false, Statement II is true .

Answer

To solve the given problem, we need to analyze the two statements regarding integrals and determine their validity.

### Statement I:
\[ \int \frac{1}{1+x^4} \, dx = \tan^{-1}(x^2) + C \]

### Statement II:
\[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C \]

### Step-by-Step Solution:

#### Step 1: Analyze Statement I
Let:
\[ I = \int \frac{1}{1+x^4} \, dx \]

To evaluate this integral, we can use a substitution or partial fraction decomposition. However, we will first check if the provided statement is correct by differentiating the right-hand side.

#### Step 2: Differentiate the Right-Hand Side of Statement I
Differentiate \(\tan^{-1}(x^2) + C\):
\[ \frac{d}{dx} \left( \tan^{-1}(x^2) \right) = \frac{2x}{1 + (x^2)^2} = \frac{2x}{1 + x^4} \]

Thus, we have:
\[ \frac{d}{dx} \left( \tan^{-1}(x^2) + C \right) = \frac{2x}{1 + x^4} \]

Since the left-hand side of Statement I is \(\frac{1}{1+x^4}\), we see that:
\[ \frac{d}{dx} \left( \tan^{-1}(x^2) + C \right) \neq \frac{1}{1+x^4} \]

This shows that Statement I is **false**.

#### Step 3: Analyze Statement II
Let:
\[ J = \int \frac{1}{1+x^2} \, dx \]

#### Step 4: Differentiate the Right-Hand Side of Statement II
Differentiate \(\tan^{-1}(x) + C\):
\[ \frac{d}{dx} \left( \tan^{-1}(x) \right) = \frac{1}{1+x^2} \]

Thus, we have:
\[ \frac{d}{dx} \left( \tan^{-1}(x) + C \right) = \frac{1}{1+x^2} \]

This confirms that Statement II is **true**.

### Conclusion:
- **Statement I** is **false**.
- **Statement II** is **true**.

### Final Answer:
- Statement I is false.
- Statement II is true.