Read the following passages and answer the following questions (7-9)
Consider the integrals of the form `l=inte^(x)(f(x)+f'(x))dx` By product rule considering `e^(x)f(x)` as first integral and `e^(x)f'(x)` as second one, we get `l=e^(x)f(x)-int(f(x)+f'(x))dx=e^(x)f(x)+c`
`int((1)/(lnx)-(1)/((lnx)^(2)))dx` is equal to

Options

A. `ln(lnx)+c`
B. `x+lnx+c`
C. `(x)/(lnx)+c`
D. none of these

Answer

To solve the integral \( I = \int \left( \frac{1}{\ln x} - \frac{1}{(\ln x)^2} \right) dx \), we can use integration by parts. Here’s a step-by-step solution:

### Step 1: Identify the components for integration by parts
We will let:
- \( u = \frac{1}{\ln x} \) (First function)
- \( dv = dx \) (Second function)

### Step 2: Differentiate and integrate
Now we differentiate \( u \) and integrate \( dv \):
- \( du = -\frac{1}{(\ln x)^2} \cdot \frac{1}{x} \, dx \) (using the chain rule)
- \( v = x \) (integrating \( dv \))

### Step 3: Apply the integration by parts formula
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ I = \left( \frac{1}{\ln x} \cdot x \right) - \int x \left( -\frac{1}{(\ln x)^2} \cdot \frac{1}{x} \right) dx \]
This simplifies to:
\[ I = \frac{x}{\ln x} + \int \frac{1}{(\ln x)^2} dx \]

### Step 4: Rearranging the equation
Now, we have:
\[ I - \int \frac{1}{(\ln x)^2} dx = \frac{x}{\ln x} \]
Let’s denote \( J = \int \frac{1}{(\ln x)^2} dx \). Therefore, we can write:
\[ I = \frac{x}{\ln x} + J \]

### Step 5: Solve for \( I \)
Now, we can express \( I \) in terms of \( J \):
\[ I - J = \frac{x}{\ln x} \]
This means:
\[ I = \frac{x}{\ln x} + J \]

### Step 6: Substitute back to find \( J \)
We can substitute \( J \) back into the equation:
\[ I = \frac{x}{\ln x} + \int \frac{1}{(\ln x)^2} dx \]
This implies that the integral \( J \) is also present in the expression for \( I \).

### Step 7: Final expression
The final expression for the integral \( I \) is:
\[ I = \frac{x}{\ln x} + C \]
where \( C \) is the constant of integration.

### Summary of the Solution
Thus, the integral \( \int \left( \frac{1}{\ln x} - \frac{1}{(\ln x)^2} \right) dx \) evaluates to:
\[ \int \left( \frac{1}{\ln x} - \frac{1}{(\ln x)^2} \right) dx = \frac{x}{\ln x} + C \]