Question

the integral of (1-3y)/ (sqrt(2y-3y^2) is ?

Answer 1

\[\int\frac{1-3y}{\sqrt{2y-3y^2}}~dy\]
Note that substituting \(u=2y-3y^2\) yields \(du=(2-6y)~dy\), but we want a substitution such that \(du=(1-3y)~dy\), so factor out a constant in the denominator:
\[\int\frac{1-3y}{\sqrt2\sqrt{y-\frac{3}{2}y^2}}~dy\]
Now, letting \(u=y-\dfrac{3}{2}y^2\) works perfectly, since \(du=(1-3y)~dy\):
\[\frac{1}{\sqrt2}\int\frac{du}{\sqrt{u}}\]