Evaluate the integral: I = integral x ln²x dx

Evaluate the integral: $$ I = \int x\ln^2x\,dx. $$

Step 1 · Set up integration by parts

Let:

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

Step 2 · Apply the formula

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

Step 3 · Reuse the known result

From Problem 143:

$$ \int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4}. $$

Substituting:

$$ I = \frac{x^2}{2}\ln^2x - \left(\frac{x^2}{2}\ln x - \frac{x^2}{4}\right). $$

Step 4 · Simplify

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

Summary Table

Additional Notes

• The domain requires $x>0$.
• Each additional power of $\ln x$ in the integrand costs one more application of integration by parts, reducing the power by one each time — a reduction-formula pattern that continues for $\int x\ln^nx\,dx$ for any $n$.
• The result can be checked by differentiating $\tfrac{x^2}{2}\ln^2x-\tfrac{x^2}{2}\ln x+\tfrac{x^2}{4}$ and confirming it returns $x\ln^2x$.