Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 419: 56

$\frac{ln(16)}{5}$

$\int_{0}^{3}\frac{dx}{5x+1}$ Let $u = 5x+1$ $\frac{du}{dx} = 5$ $dx = \frac{du}{5}$ When $x = 0$, then $u = 1$ When $x = 3$, then $u = 16$ $\int_{1}^{16}\frac{1}{5}\cdot \frac{du}{u}$ $= \frac{1}{5}~[~ln(u)\vert_{1}^{16}~]$ $= \frac{1}{5}~[~ln(16) -ln(1)~]$ $= \frac{1}{5}~[~ln(16) -0]$ $= \frac{ln(16)}{5}$