Evaluate the integral: I = integral x ln x dx

Evaluate the integral: $$ I = \int x\ln x\,dx. $$

Step 1 · Set up integration by parts

Let:

$$ u = \ln x \quad \Rightarrow \quad du = \frac{dx}{x}, \qquad dv = x\,dx \quad \Rightarrow \quad v = \frac{x^2}{2}. $$

Step 2 · Apply the formula

$$ I = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot\frac{dx}{x} = \frac{x^2}{2}\ln x - \frac12\int x\,dx. $$

Step 3 · Integrate

Since $\int x\,dx=\dfrac{x^2}{2}$:

$$ I = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C. $$

Summary Table

Additional Notes

• The domain requires $x>0$.
• This is the $n=1$ case of the general formula derived in Problem 145; setting $n=1$ there gives $\tfrac{x^2}{2}\ln x-\tfrac{x^2}{4}$, matching this result exactly.
• The result can be checked by differentiating $\tfrac{x^2}{2}\ln x-\tfrac{x^2}{4}$ and confirming it returns $x\ln x$.