Find $\displaystyle\int x e^x\,dx$.

Integration by parts: $\displaystyle\int u\,dv = uv - \int v\,du$. Choose $u = x$ (algebraic, easy to differentiate) and $dv = e^x\,dx$ (exponential, easy to integrate). Then $du = dx$ and $v = e^x$. $\displaystyle\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = (x - 1) e^x + C$. Verify by differentiating: $\dfrac{d}{dx}\left[(x-1)e^x\right] = 1 \cdot e^x + (x-1) \cdot e^x = e^x + (x-1) e^x = x e^x$ ✓. The **LIATE rule** (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) suggests which function to set as $u$. Here algebraic ($x$) beats exponential ($e^x$), so $u = x$.