$\displaystyle \int \dfrac{(8x + 1) \, dx}{\sqrt{4x - 3}}$
Let
$u = \sqrt{4x - 3}$
$u^2 = 4x - 3$
$x = \frac{1}{4}(u^2 + 3)$
$dx = \frac{1}{4}(2u \, du) = \frac{1}{2}u \, du$
Hence,
$\displaystyle \int \dfrac{(8x + 1) \, dx}{\sqrt{4x - 3}}$
$\qquad = \displaystyle \int \dfrac{[ \, 8 \cdot \frac{1}{4}(u^2 + 3) + 1 \, ] \, (\frac{1}{2}u \, du)}{u}$
$\qquad = {\displaystyle \int} (2u^2 + 7) (\frac{1}{2} \, du)$
$\qquad = {\displaystyle \int} (u^2 + \frac{7}{2}) \, du$
$\qquad = \frac{1}{3}u^3 + \frac{7}{2}u + C$
Revert to original variable of integration
$\displaystyle \int \dfrac{(8x + 1) \, dx}{\sqrt{4x - 3}}$
$\qquad = \frac{1}{3} \left( \sqrt{4x - 3} \right)^3 + \frac{7}{2}\sqrt{4x - 3} + C$
$\qquad = \frac{1}{3}(4x - 3)^{3/2} + \frac{7}{2}(4x - 3)^{1/2} + C$ ← answer
