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