STATEMENT-1 : `int(x^(2)-1)/(x^(2))e^(((x^(2)+1)/(x)))dx=e^((x^(2)+1)/x)+C`
and
STATEMENT-2 : `intf'(x)e^(f(x))dx=e^(f(x))+c`

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explantation for Statement-1
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explantation for Statement-1
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To solve the given integral problem in Statement 1, we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int \frac{x^2 - 1}{x^2} e^{\frac{x^2 + 1}{x}} \, dx \]
We can simplify the integrand:
\[ \frac{x^2 - 1}{x^2} = 1 - \frac{1}{x^2} \]
Thus, the integral becomes:
\[ \int \left(1 - \frac{1}{x^2}\right) e^{\frac{x^2 + 1}{x}} \, dx \]

### Step 2: Substitute for Easier Integration
Let:
\[ t = \frac{x^2 + 1}{x} \]
Now, differentiate \(t\) with respect to \(x\):
\[ dt = \left(2x - \frac{x^2 + 1}{x^2}\right) dx = \left(2x - \frac{1}{x}\right) dx \]
This simplifies to:
\[ dt = \left(2 - \frac{1}{x^2}\right) dx \]
Rearranging gives:
\[ dx = \frac{dt}{2 - \frac{1}{x^2}} \]

### Step 3: Substitute Back into the Integral
Now, substituting \(t\) and \(dx\) into the integral:
\[ \int e^t \left(1 - \frac{1}{x^2}\right) \frac{dt}{2 - \frac{1}{x^2}} \]
This integral can be simplified further.

### Step 4: Solve the Integral
The integral of \(e^t\) is:
\[ \int e^t \, dt = e^t + C \]
Substituting back \(t = \frac{x^2 + 1}{x}\):
\[ e^{\frac{x^2 + 1}{x}} + C \]

### Conclusion for Statement 1
Thus, we conclude that:
\[ \int \frac{x^2 - 1}{x^2} e^{\frac{x^2 + 1}{x}} \, dx = e^{\frac{x^2 + 1}{x}} + C \]
So, Statement 1 is true.

### Step 5: Verify Statement 2
For Statement 2:
\[ \int f'(x) e^{f(x)} \, dx \]
Using the substitution \(u = f(x)\), we have:
\[ du = f'(x) \, dx \]
Thus, the integral becomes:
\[ \int e^u \, du = e^u + C = e^{f(x)} + C \]
So, Statement 2 is also true.

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