Use a Table of Integrals to find $\int \dfrac{3x+5}{\sqrt{3x+6}}\,dx$.

Solution Find the subsection of the table titled Integrals Containing $\sqrt{ax+b}$ (the square root of a linear expression). Then look for an integral whose form closely resembles the problem. The closest one is an integral with $x$ in the numerator and $\sqrt{ax+b}$ in the denominator, $$\begin{equation*} \hbox{Integral 42:}\qquad\int \dfrac{x~dx}{\sqrt{a+bx}}=\dfrac{2}{3b^{2}}(bx-2a) \sqrt{a+bx}+C\tag{1} \end{equation*}$$

To express the given integral as one with a single variable in the numerator, use the substitution $u=3x+5.$ Then $du=3\,{\it dx}.$ Since $$\sqrt{3x+6}=\sqrt{(3x+5) +1}=\sqrt{u+1}$$

we find $$\int \dfrac{3x+5}{\sqrt{3x+6}}\,dx \underset{\underset{\color{#0066A7}{\hbox{\(u=3x+5, \tfrac{1}{3}du=dx\)}}}{\color{#0066A7}{\uparrow }}}{=} \dfrac{1}{3}\int \dfrac{u\,du}{\sqrt{u+1}} \\ $$

which is in the form of (1) with $a=1$ and $b=1$ So, $$\begin{align*} \int \dfrac{3x+5}{\sqrt{3x+6}}\,dx = \dfrac{1}{3}\int \dfrac{udu}{\sqrt{u+1}} \underset{\underset{\color{#0066A7}{\hbox{(1)}}}{{\color{#0066A7}\uparrow}}}{=}\dfrac{1}{3} \cdot \dfrac{2}{3} (u-2) \sqrt{1+u}+C \\ \underset{\underset{\color{#0066A7}{u=3x+5}}{\color{#0066A7}{\uparrow}}}{=}\dfrac{2}{9}[(3x+5) -2] \sqrt{1+(3x+5)}+C=\dfrac{2}{3}(x+1) \sqrt{3x+6}+C \end{align*}$$