What is $\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x$ $int_0^4 x/(x^2+9)^(1/2) dx$ ?

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What is $\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x$ $int_0^4 x/(x^2+9)^(1/2) dx$ ?

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Answer 1 · 2017-02-15T07:20:18

$\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x = 2$ $int_0^4 x/(x^2+9)^(1/2) dx = 2$

Explanation:

Note that:

$\frac{d}{d x} (x^{2} + 9) = 2 x$ $d/(dx) (x^2 + 9) = 2x$

So:

$\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x = \int_{0}^{4} \frac{1}{2} (\frac{d}{d x} (x^{2} + 9)) {(x^{2} + 9)}^{- \frac{1}{2}} d x$ $int_0^4 x/(x^2+9)^(1/2) dx = int_0^4 1/2(d/(dx)(x^2+9))(x^2+9)^(-1/2) dx$

$\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x = {[{(x^{2} + 9)}^{\frac{1}{2}}]}_{0}^{\frac{1}{2}}$ $color(white)(int_0^4 x/(x^2+9)^(1/2) dx) = [(x^2+9)^(1/2)]_0^4$
$\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x = {(4^{2} + 9)}^{\frac{1}{2}} - {(0^{2} + 9)}^{\frac{1}{2}}$ $color(white)(int_0^4 x/(x^2+9)^(1/2) dx) = (color(blue)(4)^2+9)^(1/2)-(color(blue)(0)^2+9)^(1/2)$
$\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x = {(16 + 9)}^{\frac{1}{2}} - {(9)}^{\frac{1}{2}}$ $color(white)(int_0^4 x/(x^2+9)^(1/2) dx) = (16+9)^(1/2)-(9)^(1/2)$
$\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x = 5 - 3$ $color(white)(int_0^4 x/(x^2+9)^(1/2) dx) = 5-3$
$\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x = 2$ $color(white)(int_0^4 x/(x^2+9)^(1/2) dx) = 2$

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What is $\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x$ $int_0^4 x/(x^2+9)^(1/2) dx$ ?

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Explanation:

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What is $\int_{0}^{4} \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} d x$ $int_0^4 x/(x^2+9)^(1/2) dx$ ?