To solve the integral \( \int_{0}^{1} x \log x \, dx \), we can use integration by parts. Let's go through the steps:
### Step 1: Identify the parts for integration by parts
We will use the integration by parts formula:
\[
\int u \, dv = uv - \int v \, du
\]
Let:
- \( u = \log x \) (which means \( du = \frac{1}{x} \, dx \))
- \( dv = x \, dx \) (which means \( v = \frac{x^2}{2} \))
### Step 2: Apply integration by parts
Now, we can apply the integration by parts:
\[
\int x \log x \, dx = \left( \log x \cdot \frac{x^2}{2} \right) - \int \left( \frac{x^2}{2} \cdot \frac{1}{x} \right) dx
\]
This simplifies to:
\[
\int x \log x \, dx = \frac{x^2}{2} \log x - \int \frac{x}{2} \, dx
\]
### Step 3: Calculate the remaining integral
Now we need to compute the integral \( \int \frac{x}{2} \, dx \):
\[
\int \frac{x}{2} \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4}
\]
### Step 4: Substitute back into the equation
Substituting back, we have:
\[
\int x \log x \, dx = \frac{x^2}{2} \log x - \frac{x^2}{4}
\]
### Step 5: Evaluate the definite integral from 0 to 1
Now we need to evaluate this from 0 to 1:
\[
\int_{0}^{1} x \log x \, dx = \left[ \frac{x^2}{2} \log x - \frac{x^2}{4} \right]_{0}^{1}
\]
Calculating at the upper limit \( x = 1 \):
\[
= \left( \frac{1^2}{2} \log 1 - \frac{1^2}{4} \right) = \left( 0 - \frac{1}{4} \right) = -\frac{1}{4}
\]
Calculating at the lower limit \( x = 0 \):
As \( x \to 0 \), \( \log x \to -\infty \) but \( x^2 \log x \to 0 \) (using L'Hôpital's Rule):
\[
\lim_{x \to 0} \frac{x^2 \log x}{1} = 0
\]
Thus, the contribution from the lower limit is 0.
### Final Answer
Combining both limits, we have:
\[
\int_{0}^{1} x \log x \, dx = 0 - \left( -\frac{1}{4} \right) = -\frac{1}{4}
\]
Therefore, the value of the integral \( \int_{0}^{1} x \log x \, dx \) is:
\[
\boxed{-\frac{1}{4}}
\]
