Evaluate the improper integral $displaystylein{t}_0^1{x}^{2}lnx,dx$.

Step-by-step solution

1. Choose integration by parts: let $u=lnx$ and $dv=x^{2}dx$. Then $du=\frac{1}{x}dx$ and $v=\frac{1}{3}{x}^3$.
2. Apply the formula $\int u,dv=uv-\int v,du$: $\int x^{2}lnxdx=\frac{1}{3}{x}^{3}lnx-\int \frac{1}{3}{x}^3\cdot \frac{1}{x}dx$.
3. Simplify the remaining integral: $\int \frac{1}{3}{x}^{2}dx=\frac{1}{9}{x}^3$.
4. Thus $\int x^{2}lnxdx=\frac{1}{3}{x}^{3}lnx-\frac{1}{9}{x}^3+C$.
5. Evaluate from $0$ to $1$: at $x=1$, the expression is $0-\frac{1}{9}=-\frac{1}{9}$.
6. At $x\to 0^{+}$, $\frac{1}{3}{x}^{3}lnx\to 0$ (since $x^3$ dominates $lnx$) and $\frac{1}{9}{x}^3\to 0$, so the contribution is $0$.
7. Subtracting gives $-\frac{1}{9}$.

Concept tested: Improper integral involving logarithm

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked