The value of `x gt 1` satisfying the equation
`int_(1)^(x) tlnt dt=(1)/(4)` is

Options

A. `sqrte`
B. e
C. `e^(2)`
D. `e-1`

Answer

To solve the equation

\[ \int_{1}^{x} t \ln t \, dt = \frac{1}{4} \]

for \( x > 1 \), we will use integration by parts.

### Step 1: Set up integration by parts
We will let:
- \( u = \ln t \) (which implies \( du = \frac{1}{t} dt \))
- \( dv = t \, dt \) (which implies \( v = \frac{t^2}{2} \))

### Step 2: Apply integration by parts formula
The integration by parts formula is given by:

\[ \int u \, dv = uv - \int v \, du \]

Applying this, we get:

\[ \int t \ln t \, dt = \ln t \cdot \frac{t^2}{2} - \int \frac{t^2}{2} \cdot \frac{1}{t} \, dt \]

### Step 3: Simplify the integral
The integral simplifies to:

\[ \int t \ln t \, dt = \frac{t^2}{2} \ln t - \int \frac{t^2}{2t} \, dt = \frac{t^2}{2} \ln t - \int \frac{t}{2} \, dt \]

Calculating the last integral:

\[ \int \frac{t}{2} \, dt = \frac{t^2}{4} \]

Thus, we have:

\[ \int t \ln t \, dt = \frac{t^2}{2} \ln t - \frac{t^2}{4} + C \]

### Step 4: Evaluate the definite integral
Now we evaluate the definite integral from 1 to \( x \):

\[ \int_{1}^{x} t \ln t \, dt = \left[ \frac{t^2}{2} \ln t - \frac{t^2}{4} \right]_{1}^{x} \]

Calculating the upper limit:

\[ \left( \frac{x^2}{2} \ln x - \frac{x^2}{4} \right) \]

Calculating the lower limit (at \( t = 1 \)):

\[ \left( \frac{1^2}{2} \ln 1 - \frac{1^2}{4} \right) = 0 - \frac{1}{4} = -\frac{1}{4} \]

Thus, we have:

\[ \int_{1}^{x} t \ln t \, dt = \left( \frac{x^2}{2} \ln x - \frac{x^2}{4} \right) - \left( -\frac{1}{4} \right) \]

This simplifies to:

\[ \frac{x^2}{2} \ln x - \frac{x^2}{4} + \frac{1}{4} \]

### Step 5: Set the equation equal to \( \frac{1}{4} \)
Now we set this equal to \( \frac{1}{4} \):

\[ \frac{x^2}{2} \ln x - \frac{x^2}{4} + \frac{1}{4} = \frac{1}{4} \]

Subtract \( \frac{1}{4} \) from both sides:

\[ \frac{x^2}{2} \ln x - \frac{x^2}{4} = 0 \]

### Step 6: Factor the equation
Factoring out \( \frac{x^2}{4} \):

\[ \frac{x^2}{4} \left( 2 \ln x - 1 \right) = 0 \]

Since \( x^2 \) cannot be zero for \( x > 1 \), we have:

\[ 2 \ln x - 1 = 0 \]

### Step 7: Solve for \( x \)
Solving for \( x \):

\[ 2 \ln x = 1 \implies \ln x = \frac{1}{2} \implies x = e^{1/2} = \sqrt{e} \]

### Final Answer
Thus, the value of \( x \) satisfying the equation is:

\[ x = \sqrt{e} \]