+0  
 
0
227
1
avatar+123 

the indefinite integral form of ((-1+y)/(2 sqrt(y)))^2 

 

is   (-1+y)²         1

 ∫ ———   dy= — ((y-4) Y+2 log (y)) 

     4y               8

 

am I correct? I apologize for the deformed nature of my equation. it's difficult for me to type math on my computer.

TheMathGuy  Mar 3, 2017
 #1
avatar+92805 
0

Yes that is correct except you forgot the +c

 

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This is how I would enter it

\int\frac{(-1+y)^2}{4y}\;dy

 

now open up the LaTex button on the ribbon and paste it in :)

 

\(\int\frac{(-1+y)^2}{4y}\;dy\)

 

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this might seem a bit complicated to start with if you do not know any LaTex but this forum is a great place for you to learn.

fractions are easy and very handy, so to write 2/3 as a fraction it is just

\frac{2}{3}

and now I will past this into the  LaTex box

\(\frac{2}{3}\)

Melody  Mar 3, 2017

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