Integrate? S10/√(2+√x)dx

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Integrate? S10/√(2+√x)dx

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Answer 1 · 2018-03-02T18:59:23

$I = \frac{40}{3} {(2 + \sqrt{x})}^{\frac{3}{2}} - 80 \sqrt{2 + \sqrt{x}} + C$ $I=40/3(2+sqrt(x))^(3/2)-80sqrt(2+sqrt(x))+C$

Explanation:

I assume you mean, and the S is meant as an integral sign :)

$I = \int \frac{10}{\sqrt{2 + \sqrt{x}}} d x$ $I=int10/(sqrt(2+sqrt(x)))dx$

Make a substitution $u = 2 + \sqrt{x} \Rightarrow \frac{d u}{d x} = \frac{1}{2 \sqrt{x}}$ $u=2+sqrt(x)=>(du)/dx=1/(2sqrt(x))$

$I = \int \frac{10}{\sqrt{u}} 2 \sqrt{x} d x$ $I=int10/(sqrt(u))2sqrt(x)dx$

But $u = 2 + \sqrt{x} \Rightarrow \sqrt{x} = u - 2$ $u=2+sqrt(x)=>sqrt(x)=u-2$

$I = \int \frac{10}{\sqrt{u}} 2 (u - 2) d x = 20 \int \sqrt{u} - \frac{2}{\sqrt{u}} d u$ $I=int10/(sqrt(u))2(u-2)dx=20intsqrt(u)-2/sqrt(u)du$

Integrate

$I = 20 (\frac{2}{3} u^{\frac{3}{2}} - 4 \sqrt{u}) + C$ $I=20(2/3u^(3/2)-4sqrt(u))+C$

Substitute back $u = 2 + \sqrt{x}$ $u=2+sqrt(x)$

$I = 20 (\frac{2}{3} {(2 + \sqrt{x})}^{\frac{3}{2}} - 4 \sqrt{2 + \sqrt{x}}) + C$ $I=20(2/3(2+sqrt(x))^(3/2)-4sqrt(2+sqrt(x)))+C$

$I = \frac{40}{3} {(2 + \sqrt{x})}^{\frac{3}{2}} - 80 \sqrt{2 + \sqrt{x}} + C$ $I=40/3(2+sqrt(x))^(3/2)-80sqrt(2+sqrt(x))+C$

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Integrate? S10/√(2+√x)dx

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