`int_(1)^(e)((logx)^(2))/(x)dx=?`

Options

A. `(1)/(3)`
B. `(1)/(3)e^(2)`
C. `(1)/(3)(e^(3)-1)`
D. none of these

Explanation

To solve the integral

\[ \int_{1}^{e} \frac{(\log x)^{2}}{x} \, dx, \]

we will use the substitution method.

### Step 1: Substitution
Let \( t = \log x \). Then, we differentiate both sides to find \( dx \):
\[ \frac{dt}{dx} = \frac{1}{x} \implies dx = x \, dt = e^t \, dt. \]

### Step 2: Change of Limits
Next, we need to change the limits of integration according to our substitution:
- When \( x = 1 \), \( t = \log 1 = 0 \).
- When \( x = e \), \( t = \log e = 1 \).

Thus, the limits change from \( x = 1 \) to \( x = e \) into \( t = 0 \) to \( t = 1 \).

### Step 3: Rewrite the Integral
Now we can rewrite the integral in terms of \( t \):
\[ \int_{1}^{e} \frac{(\log x)^{2}}{x} \, dx = \int_{0}^{1} t^{2} \, dt. \]

### Step 4: Integrate
Now we can integrate \( t^{2} \):
\[ \int t^{2} \, dt = \frac{t^{3}}{3} + C. \]

### Step 5: Evaluate the Integral
We will evaluate this from \( t = 0 \) to \( t = 1 \):
\[ \left[ \frac{t^{3}}{3} \right]_{0}^{1} = \frac{1^{3}}{3} - \frac{0^{3}}{3} = \frac{1}{3} - 0 = \frac{1}{3}. \]

### Final Answer
Thus, the value of the integral is

\[ \int_{1}^{e} \frac{(\log x)^{2}}{x} \, dx = \frac{1}{3}. \]

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