Here's the question you clicked on:
RolyPoly
\[\frac{dy}{dx} = \frac{x}{y^2\sqrt{1+x}}\]
\[\frac{dy}{dx} = \frac{x}{y^2\sqrt{1+x}}\]\[y^2dy = \frac{x}{\sqrt{1+x}}dx\]\[\frac{1}{3}y^3= \int \frac{u-1}{u}du = \frac{2}{3} u^{\frac{3}{2}}- 2u^{\frac{1}{2}}+C\]\[y = 2(1+x)^{\frac{3}{2}}-6(1+x)^{\frac{1}{2}}+C)^{\frac{1}{3}}\]
http://www.wolframalpha.com/input/?i=integral+x%2Fsqrt%281%2Bx%29
Did I do something wrong again?
i couldn't understand clearly what u did, but it can be simplified to 2/3*(x-2)*sqrt(x+1)
Which part you didn't understand?
then denominator is sqrt u
OMG!I typed something wrong!! \[\frac{1}{3}y^3= \int \frac{u-1}{\sqrt u}du = \frac{2}{3} u^{\frac{3}{2}}- 2u^{\frac{1}{2}}+C\]
No wonder why you couldn't understand.. That was my mistake :( Is that clear now?
yeah...u can simplify your final answer by factoring out sqrt(x+1)
That's how the answer in my book looks like :S But thanks!
ohh.. then no need. you were correct all along. welcome ^_^