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What is $\int \frac{2}{9}﻿\sqrt{4x+5}dx$?
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zavesiljid
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.
$\frac{2}{9}\left(\frac{1}{4}\int \sqrt{u}du\right)$
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 $\frac{2}{3}{u}^{\frac{3}{2}}$
$\frac{1}{18}\left(\frac{2}{3}{u}^{\frac{3}{2}}+C\right)$
Simplify
$\frac{1}{27}{u}^{\frac{3}{2}}+C$
Replace all occurrences of u with 4x+5.
$\frac{1}{27}\left(4x+5{\right)}^{\frac{3}{2}}+C$
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