MHB How would you approach this integral?

[Lorena_Santoro]

 

[mrtwhs]

$u=x-3$

 

[Kansas Boy]

Let u= x- 3. Then x= u+ 3 and dx= du so the given integral is equal to $\int (u+ 3)u^{1/2}du= \int u^{3/2}+ 3u^{1/2}= \frac{2}{5}u^{5/2}+ 2u^{3/2}+ C= \frac{2}{5}(x- 3)^{5/2}+ 2(x- 3)^{3/2}+ C$.