The value of the integral `int_1^e(logx)^3` dx is

Options

A. `6+2e`
B. `6-2e`
C. `2e-6`
D. none of these

Answer

To solve the integral \( \int_1^e (\log x)^3 \, dx \), we will use integration by parts.

### Step-by-Step Solution:

1. **Identify the parts for integration by parts**:
We choose:
- \( u = (\log x)^3 \) (which we will differentiate)
- \( dv = dx \) (which we will integrate)

2. **Differentiate and integrate**:
- Differentiate \( u \):
\[ du = 3(\log x)^2 \cdot \frac{1}{x} \, dx = \frac{3(\log x)^2}{x} \, dx \]
- Integrate \( dv \):
\[ v = x \]

3. **Apply the integration by parts formula**:
The formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ \int (\log x)^3 \, dx = x(\log x)^3 - \int x \cdot \frac{3(\log x)^2}{x} \, dx \]
Simplifying the integral:
\[ = x(\log x)^3 - 3 \int (\log x)^2 \, dx \]

4. **Now we need to evaluate \( \int (\log x)^2 \, dx \)** using integration by parts again:
- Let \( u = (\log x)^2 \) and \( dv = dx \).
- Then, \( du = 2 \log x \cdot \frac{1}{x} \, dx = \frac{2 \log x}{x} \, dx \) and \( v = x \).
- Applying integration by parts:
\[ \int (\log x)^2 \, dx = x(\log x)^2 - \int x \cdot \frac{2 \log x}{x} \, dx \]
Simplifying:
\[ = x(\log x)^2 - 2 \int \log x \, dx \]

5. **Now we need to evaluate \( \int \log x \, dx \)**:
- Let \( u = \log x \) and \( dv = dx \).
- Then, \( du = \frac{1}{x} \, dx \) and \( v = x \).
- Applying integration by parts:
\[ \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx = x \log x - x \]

6. **Putting it all together**:
Now substituting back:
\[ \int (\log x)^2 \, dx = x(\log x)^2 - 2(x \log x - x) = x(\log x)^2 - 2x \log x + 2x \]

7. **Substituting back into the integral of \( (\log x)^3 \)**:
\[ \int (\log x)^3 \, dx = x(\log x)^3 - 3\left[x(\log x)^2 - 2x \log x + 2x\right] \]
Simplifying:
\[ = x(\log x)^3 - 3x(\log x)^2 + 6x \log x - 6x \]

8. **Evaluate from 1 to \( e \)**:
\[ \int_1^e (\log x)^3 \, dx = \left[ x(\log x)^3 - 3x(\log x)^2 + 6x \log x - 6x \right]_1^e \]

- **Upper limit \( x = e \)**:
\[ = e(1)^3 - 3e(1)^2 + 6e(1) - 6e = e - 3e + 6e - 6e = -2e \]

- **Lower limit \( x = 1 \)**:
\[ = 1(0)^3 - 3(1)(0)^2 + 6(1)(0) - 6(1) = -6 \]

Now, combining:
\[ = (-2e) - (-6) = -2e + 6 \]

### Final Result:
The value of the integral \( \int_1^e (\log x)^3 \, dx \) is:
\[ 6 - 2e \]