Question

integral of dy/sqrt(y)(1-sqrt(y))

Answer 1

\[\int\limits \frac{ dy }{ \sqrt{y}(1-\sqrt{y}) }\]

Answer 2

let u=1-sqrt(y)

Answer 3

how do i solve for du in terms of dy then?

Answer 4

are you asking how to differentiate sqrt(y)?

Answer 5

no

Answer 6

you differentiate both sides

Answer 7

try

Answer 8

i need to have du in the equation to solve the integral

Answer 9

you will end getting something like du=f(y) dy

Answer 10

so then it equals \[\int\limits \frac{ dy }{ u+1(u) }\]

Answer 11

I picked sqrt(y)=u it seemed to work pretty well for me.

Answer 12

\[\int\limits\limits \frac{ dy }{ \sqrt{y}(1-\sqrt{y}) } \\ \int\limits_{}^{} \frac{-2}{1-\sqrt{y}} \cdot \frac{1}{-2 \sqrt{y}} dy\]

Answer 13

\[\frac{ du }{ dy }=\frac{ 1 }{ 2\sqrt{y} }\]

Answer 14

with a negtaive

Answer 15

if you choose u=1-sqrt(y)

Answer 16

\[u=1-\sqrt{y} \\ du=-\frac{1}{2 \sqrt{y}} dy\]

Answer 17

\[\LARGE \int\limits \frac{dy}{\sqrt{y}-y}\] \[\LARGE \sqrt{y}=u \\ \LARGE \frac{dy}{2 \sqrt{y}}=du \\ \LARGE dy=2udu\] \[\LARGE \int\limits \frac{2u du}{u-u^2}=-2 \int\limits \frac{du}{u-1}\\ \LARGE -2\ln|u-1|+C = \ln \frac{1}{(\sqrt{y}-1)^2}+C\]

Answer 18

well then \[-2\ln \left| 1-\sqrt{y} \right|+C \] is my answer