+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.
+118680 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
\(\frac{2}{3}\)
