help indefinite integral....

[khenjin017]

indefinite integral of dy/(1+y)sqrt(y)

the first thing i thought in my mind ,this will leads to inverse formula...
so i multiply sqrt(y) to dividend and divisor.... sqrt(y)dy/(y+y^2).... but the tangent is dy/a^2+u^2...... so i try to use completing the square in divisor (y^2+y+1/4)-1/4....which is i cant use again...

next thing is i try to combine the divisor which is (y^3/2+y^1/2)... and try to use it as the 'u' ... but i cant...

so please help me ,what is way to solve this problem/..... easy but calculus is fooling me..

 

[Ishuda]

Depending on what you are familiar with, try u = y^1/2

 

[khenjin017]

i think i already get the the answer...i made the given look like this (y^-1/2)dy/1+y.. i use sec inverse formula.... so a=1
u=y^1/2 ...differentiate the u , then 2du=y^-1/2 dy... so its arc Sec sqrt(y) + c//

 

[Deleted member 4993]

i think i already get the the answer...i made the given look like this (y^-1/2)dy/1+y.. i use sec inverse formula.... so a=1
u=y^1/2 ...differentiate the u , then 2du=y^-1/2 dy... so its arc Sec sqrt(y) + c//

Differentiate your answer - do you get back the original function?

 

[khenjin017]

the final answer is 2 arc Sec sqrt(y) + c ... and by differentiating it ,yes it comes back to the given function..

 

[Deleted member 4993]

indefinite integral of dy/{(1+y)sqrt(y)}

$\displaystyle \displaystyle{\int \frac{dy}{(1+y)\sqrt{y}}}$ ← u = √y → 2 * u du = dy

$\displaystyle \displaystyle{= 2 \int \frac{du}{(1+u^2)}}$

$\displaystyle \displaystyle{= 2*tan^{-1}(\sqrt{y}) + C}$

 

[khenjin017]

oh...thanks....

over confidence to my answers is my really my problem...