If `int(log_(x)e.log_(ex)e.log_(e^(2)x)e)/(x)dx=A log_(e)(log_(e)x)+Blog_(e)(1+log_(e)x)+Clog_(e)(2+log_(e)x)+lambda,` then

Options

A. `A+B=0`
B. `A+B=-1//2`
C. `A+C=0`
D. `A-C=0`

Answer

To solve the given integral

\[ I = \int \frac{\log_e x \cdot \log_e e \cdot \log_e (e^2 x)}{x} \, dx, \]

we will first simplify the expression inside the integral.

### Step 1: Simplifying the Logarithm Expressions

Using the properties of logarithms, we can rewrite:

1. \(\log_e e = 1\)
2. \(\log_e (e^2 x) = \log_e (e^2) + \log_e x = 2 + \log_e x\)

Thus, the integral becomes:

\[ I = \int \frac{\log_e x \cdot 1 \cdot (2 + \log_e x)}{x} \, dx = \int \frac{\log_e x (2 + \log_e x)}{x} \, dx. \]

### Step 2: Expanding the Integral

Now, we can expand the integrand:

\[ I = \int \frac{2 \log_e x}{x} \, dx + \int \frac{(\log_e x)^2}{x} \, dx. \]

### Step 3: Substituting for Integration

Let \(t = \log_e x\). Then, \(dx = e^t \, dt\) and \(x = e^t\). Thus, \(\frac{1}{x} = e^{-t}\).

Now, substituting into the integral:

1. For the first integral:
\[ \int \frac{2 \log_e x}{x} \, dx = \int 2t \cdot e^{-t} \cdot e^t \, dt = \int 2t \, dt = t^2 = (\log_e x)^2. \]

2. For the second integral:
\[ \int \frac{(\log_e x)^2}{x} \, dx = \int t^2 \cdot e^{-t} \cdot e^t \, dt = \int t^2 \, dt = \frac{t^3}{3} = \frac{(\log_e x)^3}{3}. \]

### Step 4: Combining the Results

Combining both parts, we have:

\[ I = (\log_e x)^2 + \frac{(\log_e x)^3}{3} + C, \]

where \(C\) is the constant of integration.

### Step 5: Final Expression

Thus, we can express the integral as:

\[ I = \frac{1}{3} (\log_e x)^3 + (\log_e x)^2 + C. \]

### Step 6: Comparing with Given Expression

The given expression is:

\[ A \log_e (\log_e x) + B \log_e (1 + \log_e x) + C \log_e (2 + \log_e x) + \lambda. \]

By comparing coefficients, we can find the values of \(A\), \(B\), and \(C\).

### Conclusion

From the integration, we find:

- \(A = \frac{1}{3}\)
- \(B = 1\)
- \(C = 0\)