sd
Actually it should be:
$\displaystyle \frac{yy^{\prime}}{2y + 6} = \frac{1}{x}$
$\displaystyle \frac{y}{2y + 6} dy = \frac{1}{x}dx$
$\displaystyle \int \frac{y}{2y + 6} dy = \int \frac{1}{x}dx$
The RHS is
$\displaystyle \int \frac{y}{2y + 6} dy = \frac{1}{2} \int \frac{y}{y + 3} dy = \frac{1}{2} \int \left (1 - \frac{3}{y + 3} \right ) dy $
Can you do it now?
-Dan
Let me give you one more hint. Look at the RHS:
$\displaystyle \int \frac{1}{x} dx$
I presume that this integral is giving you no problems.
Is there any substitution you can think of that will make
$\displaystyle \int \frac{1}{y + 3}dy$
look similar to the x integral?
-Dan
[quote=Krizalid;60174]$\displaystyle \int {\frac{1}
{{y + 3}}~dy} = \int {\frac{{(y + 3)'}}
{{y + 3}}~dy} = \ln \left| {y + 3} \right| + k~\blacksquare$
i understand how to do this, this is simple intergration but dont understand how to relate it to 1/2 intergral (1-3/(y+3) dy