5.5 The Substitution Rule/54

$\int _{0}^{7}{\sqrt {4+3x}}\,dx$

${\begin{aligned}u&=4+3x\\[2ex]du&=3\,dx\\[2ex]{\frac {1}{3}}du&=dx\\[2ex]\end{aligned}}$

${\begin{aligned}\int _{0}^{7}{\sqrt {4+3x}}\,dx=\int _{4}^{25}{\sqrt {u}}du\\[2ex]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\end{aligned}}$