\(\int {x \sqrt{2x+1} } \, dx\) We will use u-substitution: u=2x +1 , du = 2dx,\(dx= \frac{du}{2}, x= \frac{u-1}{2}\) ...=\(\frac{1}{4} \int {(u-1)} \sqrt{u} \, du = \frac{1}{4} \int {u^{3/2} -u^{1/2} } \, du\)=
=\(\frac{1}{4} ( \frac{u^{5/2} }{ \frac{5}{2} }- \frac{u^{3/2} }{ \frac{3}{2} } ) =\)=\(\frac{1}{2} \frac{3u^{5/2} -5u^{3/2} }{15}=\)
=\(\frac{1}{30} u^{3/2} (3u-5)\)=
=\(\frac{1}{30} (2x+1)^{3/2}(6x-2)=\)
=\(\frac{1}{15} (2x+1)^{2/3}(3x-1) +C\)
