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**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