`int(lnx)^(-1)dx-int(lnx)^(-2)` dx is equal to

Options

A. `x(lnx)^(-1)+c`
B. `x(lnx)^(-2)+c`
C. `x(lnx)+c`
D. `x(lnx)^(2)+c`

Answer

To solve the problem \( \int (\ln x)^{-1} \, dx - \int (\ln x)^{-2} \, dx \), we will use integration by parts. Let's break it down step by step.

### Step 1: Set up the integration by parts
We start with the first integral \( \int (\ln x)^{-1} \, dx \). We will use integration by parts, which states:

\[ \int u \, dv = uv - \int v \, du \]

Let:
- \( u = (\ln x)^{-1} \)
- \( dv = dx \)

### Step 2: Differentiate and integrate
Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = -(\ln x)^{-2} \cdot \frac{1}{x} \, dx \]
- Integrate \( dv \):
\[ v = x \]

### Step 3: Apply integration by parts
Substituting into the integration by parts formula, we have:
\[ \int (\ln x)^{-1} \, dx = x (\ln x)^{-1} - \int x \left(-(\ln x)^{-2} \cdot \frac{1}{x}\right) \, dx \]
This simplifies to:
\[ \int (\ln x)^{-1} \, dx = x (\ln x)^{-1} + \int (\ln x)^{-2} \, dx \]

### Step 4: Rearranging the equation
Now we can rearrange the equation:
\[ \int (\ln x)^{-1} \, dx - \int (\ln x)^{-2} \, dx = x (\ln x)^{-1} + C \]

### Step 5: Final expression
Thus, we have:
\[ \int (\ln x)^{-1} \, dx - \int (\ln x)^{-2} \, dx = x (\ln x)^{-1} + C \]

### Conclusion
The final result is:
\[ \int (\ln x)^{-1} \, dx - \int (\ln x)^{-2} \, dx = x (\ln x)^{-1} + C \]