Question

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

Answer 1

\[\frac{dy}{dx} = \frac{x}{y^2\sqrt{1+x}}\]\[y^2dy = \frac{x}{\sqrt{1+x}}dx\]\[\frac{1}{3}y^3= \int \frac{u-1}{u}du = \frac{2}{3} u^{\frac{3}{2}}- 2u^{\frac{1}{2}}+C\]\[y = 2(1+x)^{\frac{3}{2}}-6(1+x)^{\frac{1}{2}}+C)^{\frac{1}{3}}\]

Answer 2

http://www.wolframalpha.com/input/?i=integral+x%2Fsqrt%281%2Bx%29

Answer 3

Did I do something wrong again?

Answer 4

i couldn't understand clearly what u did, but it can be simplified to 2/3*(x-2)*sqrt(x+1)

Answer 5

Which part you didn't understand?

Answer 6

u=?

Answer 7

1+x

Answer 8

then denominator is sqrt u

Answer 9

OMG!I typed something wrong!!
\[\frac{1}{3}y^3= \int \frac{u-1}{\sqrt u}du = \frac{2}{3} u^{\frac{3}{2}}- 2u^{\frac{1}{2}}+C\]

Answer 10

No wonder why you couldn't understand.. That was my mistake :(
Is that clear now?

Answer 11

yeah...u can simplify your final answer by factoring out sqrt(x+1)

Answer 12

if needed.

Answer 13

That's how the answer in my book looks like :S
But thanks!

Answer 14

ohh.. then no need. you were correct all along.
welcome ^_^