Answer
a)$\frac{2}{5}\sqrt{5x+8}+C$
b)$\frac{2}{5}\sqrt{5x+8}+C$
Work Step by Step
a)Let $u=5x+8=>du=5 dx=>\frac{1}{5}du=dx$
$\int \frac{dx}{\sqrt{5x+8}} =\int \frac{1}{5}(\frac{1}{\sqrt{u}})du$
=$\frac{1}{5} \int u^{-1/2}du$
=$\frac{1}{5}(2u^{1/2})+C$
=$\frac{2}{5}\sqrt{5x+8}+C$
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b)Let u=$\sqrt{5x+8} =>du=\frac{1}{2}(5x+8)^\frac{-1}{2}$(5)dx
=$\frac{2}{5}du=\frac{dx}{\sqrt{5x+8}}$
=$\int \frac{dx}{\sqrt{5x+8}}=\int \frac{2}{5}du$
=$\frac{2}{5}u+C$
=$\frac{2}{5}\sqrt{5x+8}+C$
