The value of integral `int_(1)^(e) (log x)^(3)dx` , is

Options

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

Answer

To solve the integral \( I = \int_{1}^{e} (\log x)^3 \, dx \), we will use a substitution method. Here are the steps to find the value of the integral:

### Step 1: Substitution
Let \( x = e^y \). Then, the differential \( dx \) can be expressed as:
\[ dx = e^y \, dy \]

### Step 2: Change the limits
When \( x = 1 \):
\[ y = \log(1) = 0 \]
When \( x = e \):
\[ y = \log(e) = 1 \]
Thus, the limits change from \( x = 1 \) to \( x = e \) into \( y = 0 \) to \( y = 1 \).

### Step 3: Rewrite the integral
Substituting \( x \) and \( dx \) into the integral, we have:
\[ I = \int_{0}^{1} (\log(e^y))^3 e^y \, dy \]
Using the property of logarithms, \( \log(e^y) = y \):
\[ I = \int_{0}^{1} (y)^3 e^y \, dy \]

### Step 4: Integration by parts
We will use integration by parts to solve \( \int y^3 e^y \, dy \). Let:
- \( u = y^3 \) → \( du = 3y^2 \, dy \)
- \( dv = e^y \, dy \) → \( v = e^y \)

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int y^3 e^y \, dy = y^3 e^y - \int e^y (3y^2) \, dy \]

### Step 5: Repeat integration by parts
Now we need to evaluate \( \int 3y^2 e^y \, dy \) using integration by parts again:
- Let \( u = 3y^2 \) → \( du = 6y \, dy \)
- \( dv = e^y \, dy \) → \( v = e^y \)

Thus,
\[ \int 3y^2 e^y \, dy = 3y^2 e^y - \int e^y (6y) \, dy \]

### Step 6: Repeat again
Now we evaluate \( \int 6y e^y \, dy \):
- Let \( u = 6y \) → \( du = 6 \, dy \)
- \( dv = e^y \, dy \) → \( v = e^y \)

Thus,
\[ \int 6y e^y \, dy = 6y e^y - \int 6 e^y \, dy = 6y e^y - 6 e^y \]

### Step 7: Combine results
Now, substituting back:
\[ \int y^3 e^y \, dy = y^3 e^y - (3y^2 e^y - (6y e^y - 6 e^y)) \]
Combining everything:
\[ = y^3 e^y - 3y^2 e^y + 6y e^y - 6 e^y \]

### Step 8: Evaluate from 0 to 1
Now we evaluate the expression from 0 to 1:
\[ = \left[ y^3 e^y - 3y^2 e^y + 6y e^y - 6 e^y \right]_{0}^{1} \]
Calculating at \( y = 1 \):
\[ = (1^3 e^1 - 3(1^2)e^1 + 6(1)e^1 - 6e^1) = (e - 3e + 6e - 6e) = -2e \]
Calculating at \( y = 0 \):
\[ = (0 - 0 + 0 - 6) = -6 \]
Thus, the final value is:
\[ I = (-2e + 6) - 0 = 6 - 2e \]

### Final Answer
The value of the integral is:
\[ \int_{1}^{e} (\log x)^3 \, dx = 6 - 2e \]