MHB How would you approach this integral?

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

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