Answer
See proof
Work Step by Step
$$\eqalign{
& \int {x\sqrt {a + bx} } dx = \frac{2}{{15{b^2}}}\left( {3bx - 2a} \right){\left( {a + bx} \right)^{3/2}} + C \cr
& {\text{Let }}F\left( x \right) = \frac{2}{{15{b^2}}}\left( {3bx - 2a} \right){\left( {a + bx} \right)^{3/2}} + C\cr\
&{\text{We will show that F is an antiderivative of }} \cr
& x\sqrt {a + bx} \cr
& {\text{Differentiating }}{\frac{2}{{15{b^2}}}\left( {3bx - 2a} \right){{\left( {a + bx} \right)}^{3/2}} + C} \cr
& \frac{2}{{15{b^2}}}\underbrace {\frac{d}{{dx}}\left[ {\left( {3bx - 2a} \right){{\left( {a + bx} \right)}^{3/2}}} \right]}_{{\text{Use the product rule}}} + \frac{d}{{dx}}\left[ C \right] \cr
&= \frac{2}{{15{b^2}}}\left( {3bx - 2a} \right)\left( {\frac{3}{2}} \right){\left( {a + bx} \right)^{1/2}}\left( b \right) + \frac{2}{{15{b^2}}}{\left( {a + bx} \right)^{3/2}}\left( {3b} \right) \cr
& {\text{Simplifying}} \cr
& =\frac{1}{{5b}}\left( {3bx - 2a} \right){\left( {a + bx} \right)^{1/2}} + \frac{2}{{5b}}{\left( {a + bx} \right)^{3/2}} \cr
& {\text{Factoring}} \cr
& =\frac{1}{{5b}}{\left( {a + bx} \right)^{1/2}}\left( {3bx - 2a + 2\left( {a + bx} \right)} \right) \cr
& =\frac{1}{{5b}}{\left( {a + bx} \right)^{1/2}}\left( {3bx - 2a + 2a + 2bx} \right) \cr
& =\frac{1}{{5b}}\sqrt {a + bx} \left( {5bx} \right)\cr
& = x\sqrt {a + bx} \cr
& {\text{The statement is true}}{\text{, then the formula is correct}}{\text{.}} \cr} $$
