Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 4

Check by differentiation $\displaystyle\frac{x}{\sqrt{a+bx}} dx = \frac{2}{3b^2} (bx - 2a) \sqrt{a + bx} + C$
We take the derivative of the expression to the right integral

$
\begin{equation}
\begin{aligned}
f(x) &=\frac{2}{3b^2} (bx - 2a) \sqrt{a+bx} + C \\
\\
f(x) &= \left( \frac{3bx}{3b^2} - \frac{4a}{3b^2} \right) \sqrt{a+bx} + C\\
\\
f(x) &= \left( \frac{2x}{3b} - \frac{4a}{3b^2} \right) (a + bx)^{\frac{1}{2}} + C \\
\\
f'(x) &= \left[ \left( \frac{2x}{3b} - \frac{4a}{3b^2} \right) \frac{d}{dx} (a + bx)^{\frac{1}{2}} + (a + bx)^{\frac{1}{2}} \frac{d}{dx} \left( \frac{2x}{3b} - \frac{4a}{3b^2} \right) \right] + \frac{d}{dx} C\\
\\
f'(x) &= \left( \frac{2x}{3b} - \frac{4a}{3b^2} \right) \left( \frac{1}{2} \right) (a + bx)^{\frac{-1}{2}} \frac{d}{dx} ( a + bx) + (a + bx)^{\frac{1}{2}} \left( \frac{2}{3b} - 0 \right) + 0 \\
\\
f'(x) &= \left( \frac{2x}{3b} - \frac{4a}{3b^2} \right) \left( \frac{1}{2} \right) (a + bx)^{\frac{-1}{2}} (b)+ (a+bx)^{\frac{1}{2}} \left( \frac{2}{3b} \right)\\
\\
f'(x) &= \frac{\left(\frac{2x}{3b} - \frac{4a}{3b^2} \right)\left( \frac{b}{2} \right)}{(a+bx)^{\frac{1}{2}}} + (a + bx)^{\frac{1}{2}} \left( \frac{2}{3b} \right)\\
\\
f'(x) &= \frac{\frac{2 \cancel{b}x}{6\cancel{b}} - \frac{4ab}{3b^2} }{(a + bx)^{\frac{1}{2}}} + (a + bx)^{\frac{1}{2}} \left( \frac{2}{3b} \right) \\
\\
f'(x) &= \frac{\frac{x}{3} - \cancel{\frac{2a}{3b}} + \cancel{\frac{2a}{3b}} + \frac{2 \cancel{b}x}{3 \cancel{b}} }{(a + bx)^{\frac{1}{2}}}\\
\\
f'(x) &= \frac{\frac{x}{3}+\frac{2x}{3} }{(a+bx)^{\frac{1}{2}}}\\
\\
f'(x) &= \frac{\frac{x+2x}{3}}{(a+bx)^{\frac{1}{2}}}\\
\\
f'(x) &= \frac{\frac{\cancel{3}x}{\cancel{3}}}{(a+bx)^{\frac{1}{2}}}\\
\\
f'(x) &= \frac{x}{(a+bx)^{\frac{1}{2}}} \qquad \text{ or } \qquad f'(x) = \frac{x}{\sqrt{a+bx}}
\end{aligned}
\end{equation}
$


The expression to the left is correct