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
Let:
- \( u = \ln t \) ⇒ \( du = \frac{1}{t} dt \)
- \( dv = t dt \) ⇒ \( v = \frac{t^2}{2} \)

Using the integration by parts formula:

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

we can express our integral as:

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

### Step 2: Simplify the integral
Now, simplifying the second integral:

\[ \int \frac{t^2}{2} \cdot \frac{1}{t} \, dt = \int \frac{t}{2} \, dt = \frac{1}{2} \cdot \frac{t^2}{2} = \frac{t^2}{4} \]

Thus, we have:

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

### Step 3: 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 this gives:

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

Since \( \ln 1 = 0 \):

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

### Step 4: Set the equation equal to \(\frac{1}{4}\)
Setting this equal to \(\frac{1}{4}\):

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

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

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

### Step 5: Factor out common terms
Factoring out \(\frac{x^2}{2}\):

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

This gives us two cases:

1. \( \frac{x^2}{2} = 0 \) (not possible since \( x > 1 \))
2. \( \ln x - \frac{1}{2} = 0 \)

### Step 6: Solve for \( x \)
From the second case:

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

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

\[ \boxed{\sqrt{e}} \]