Evaluate ∫₀³ x/(√(x² + 16)) dx with the substitution u = x² + 16…

Step 1. New limits. At $x = 0$, $u = 16$. At $x = 3$, $u = 25$.

Step 2. Substitute. $x\,dx = \tfrac{1}{2}du$, so the integral is $\tfrac{1}{2}\displaystyle\int_{16}^{25}u^{-1/2}\,du = \tfrac{1}{2}\left[2u^{1/2}\right]_{16}^{25}$.

Step 3. Evaluate. $\left[\sqrt{u}\right]_{16}^{25} = 5 - 4 = 1$.