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Since $$\frac{2}{9}$$ is constant with respect to x, move $$\frac{2}{9}$$ out of the integral.
$$\frac{2}{9}\int \sqrt{4x+5}dx$$
Let u=4x+5. Then du=4dx, so $$\frac{1}{4}du=dx$$. Rewrite using u and du.
$$\frac{2}{9}\int \sqrt{u}\frac{1}{4}du$$
Combine $$\sqrt{u}$$ and $$\frac{1}{4}$$.
$$\frac{2}{9}\int \frac{\sqrt{u}}{4}du$$
Since $$\frac{1}{4}$$ is constant with respect to u, move $$\frac{1}{4}$$ out of the integral.
$$\frac{2}{9}(\frac{1}{4}\int \sqrt{u}du)$$
Simplify the expression.
$$\frac{1}{18}\int u^{\frac{1}{2}}du$$
By the Power Rule, the integral of $$u^{\frac{1}{2}}$$ with respect to u is $$\frac{2}{3}{u}^{\frac{3}{2}}$$
$$\frac{1}{18}(\frac{2}{3}{u}^{\frac{3}{2}}+C)$$
Simplify
$$\frac{1}{27}{u}^{\frac{3}{2}}+C$$
Replace all occurrences of u with 4x+5.
$$\frac{1}{27}(4x+5{)}^{\frac{3}{2}}+C$$