What is $\int \frac{x}{{(x + 3)}^{\frac{1}{2}}} d x$ $int x / (x+3)^(1/2) dx$ ?

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What is $\int \frac{x}{{(x + 3)}^{\frac{1}{2}}} d x$ $int x / (x+3)^(1/2) dx$ ?

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Answer 1 · 2018-03-11T21:54:53

$\int \frac{x}{{(x + 3)}^{\frac{1}{2}}} d x = \frac{2}{3} {(x + 3)}^{\frac{1}{2}} (x - 6) + C$ $int x/(x+3)^(1/2) dx = 2/3(x+3)^(1/2)(x-6)+C$

Explanation:

$\int \frac{x}{{(x + 3)}^{\frac{1}{2}}} d x = \int \frac{(x + 3) - 3}{{(x + 3)}^{\frac{1}{2}}} d x$ $int x/(x+3)^(1/2) dx = int ((x+3)-3)/(x+3)^(1/2) dx$

$\int \frac{x}{{(x + 3)}^{\frac{1}{2}}} d x = \int {(x + 3)}^{\frac{1}{2}} - 3 {(x + 3)}^{- \frac{1}{2}} d x$ $color(white)(int x/(x+3)^(1/2) dx) = int (x+3)^(1/2)-3(x+3)^(-1/2) dx$

$\int \frac{x}{{(x + 3)}^{\frac{1}{2}}} d x = \frac{2}{3} {(x + 3)}^{\frac{3}{2}} - 6 {(x + 3)}^{\frac{1}{2}} + C$ $color(white)(int x/(x+3)^(1/2) dx) = 2/3(x+3)^(3/2)-6(x+3)^(1/2)+C$

$\int \frac{x}{{(x + 3)}^{\frac{1}{2}}} d x = \frac{2}{3} {(x + 3)}^{\frac{1}{2}} ((x + 3) - 9) + C$ $color(white)(int x/(x+3)^(1/2) dx) = 2/3(x+3)^(1/2)((x+3)-9)+C$

$\int \frac{x}{{(x + 3)}^{\frac{1}{2}}} d x = \frac{2}{3} {(x + 3)}^{\frac{1}{2}} (x - 6) + C$ $color(white)(int x/(x+3)^(1/2) dx) = 2/3(x+3)^(1/2)(x-6)+C$

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What is $\int \frac{x}{{(x + 3)}^{\frac{1}{2}}} d x$ $int x / (x+3)^(1/2) dx$ ?

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