`int(logx)/x^2dx =`

Options

A. `1/x(logx+1)`
B. `-1/x (log x+1)`
C. `1/x (log x-1)`
D. `log (x+1)`

Answer

To solve the integral \( \int \frac{\log x}{x^2} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = \log x \). Then, the derivative of \( t \) with respect to \( x \) is given by:
\[ dt = \frac{1}{x} \, dx \quad \Rightarrow \quad dx = x \, dt \]
Since \( x = e^t \), we can substitute \( dx \) in terms of \( t \):
\[ dx = e^t \, dt \]

### Step 2: Rewrite the Integral
Now, we can rewrite the integral using the substitution:
\[ \int \frac{\log x}{x^2} \, dx = \int \frac{t}{(e^t)^2} \cdot e^t \, dt = \int \frac{t}{e^t} \, dt \]

### Step 3: Integration by Parts
To solve \( \int \frac{t}{e^t} \, dt \), we will use integration by parts. Let:
- \( u = t \) (thus \( du = dt \))
- \( dv = e^{-t} \, dt \) (thus \( v = -e^{-t} \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int \frac{t}{e^t} \, dt = -t e^{-t} - \int -e^{-t} \, dt \]
\[ = -t e^{-t} + \int e^{-t} \, dt \]
\[ = -t e^{-t} - e^{-t} + C \]

### Step 4: Factor Out Common Terms
Now, we can factor out \( -e^{-t} \):
\[ = -e^{-t}(t + 1) + C \]

### Step 5: Substitute Back
Now we substitute back \( t = \log x \):
\[ = -\frac{1}{e^{\log x}}(\log x + 1) + C \]
Since \( e^{\log x} = x \):
\[ = -\frac{\log x + 1}{x} + C \]

### Final Answer
Thus, the final answer for the integral \( \int \frac{\log x}{x^2} \, dx \) is:
\[ -\frac{\log x + 1}{x} + C \]