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RolyPoly
 3 years ago
\[\frac{dy}{dx} = \frac{x}{y^2\sqrt{1+x}}\]
RolyPoly
 3 years ago
\[\frac{dy}{dx} = \frac{x}{y^2\sqrt{1+x}}\]

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RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dx} = \frac{x}{y^2\sqrt{1+x}}\]\[y^2dy = \frac{x}{\sqrt{1+x}}dx\]\[\frac{1}{3}y^3= \int \frac{u1}{u}du = \frac{2}{3} u^{\frac{3}{2}} 2u^{\frac{1}{2}}+C\]\[y = 2(1+x)^{\frac{3}{2}}6(1+x)^{\frac{1}{2}}+C)^{\frac{1}{3}}\]

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=integral+x%2Fsqrt%281%2Bx%29

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0Did I do something wrong again?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0i couldn't understand clearly what u did, but it can be simplified to 2/3*(x2)*sqrt(x+1)

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0Which part you didn't understand?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0then denominator is sqrt u

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0OMG!I typed something wrong!! \[\frac{1}{3}y^3= \int \frac{u1}{\sqrt u}du = \frac{2}{3} u^{\frac{3}{2}} 2u^{\frac{1}{2}}+C\]

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0No wonder why you couldn't understand.. That was my mistake :( Is that clear now?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0yeah...u can simplify your final answer by factoring out sqrt(x+1)

RolyPoly
 3 years ago
Best ResponseYou've already chosen the best response.0That's how the answer in my book looks like :S But thanks!

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0ohh.. then no need. you were correct all along. welcome ^_^
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