Evaluate $\displaystyle\int \dfrac{8x + 13}{\sqrt{4x + 7}} \; dx$
Let $I = \displaystyle\int \dfrac{8x + 3}{\sqrt{4x + 7}} \; dx$
Let $4x + 7 = u$ $\;\;\; \cdots$ (1) $\implies$ $x = \dfrac{u - 7}{4}$
Differentiating equation (1) gives
$4 dx = du$ $\implies$ $dx = \dfrac{du}{4}$
Now, $8x + 13 = 8 \left(\dfrac{u - 7}{4}\right) + 13 = 2 u - 1$
$\begin{aligned}
\therefore \; I & = \dfrac{1}{4} \int \dfrac{2 u - 1}{\sqrt{u}} \; du \\\\
& = \dfrac{1}{2} \times u^{3/2} \times \dfrac{2}{3} - \dfrac{1}{4} \times u^{1/2} \times 2 + c \\\\
& = \dfrac{1}{3} \left(4x + 7\right)^{3/2} - \dfrac{1}{2} \left(4x + 7\right)^{1/2} + c
\end{aligned}$