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3x²+5y+5x(dy/dx)-4y(dy/dx)=0

3x²+5y=(5x-4y)dy/dx

dy/dx=(3x²+5y)/(5x-4y)

Using the quotient rule: Let u=x and v=2x+5 then du/dx= 1 and dv/dx=2 Hence dy/dx=[(2x+5)x1- (2x)]/(2x+5)²             =5/(2x+5)²

Integrate by parts: integral = [uv] - ∫u'v dx (u'= derivative of u, v'= derivative of v)

u= x     u'= 1

v' = sin2x        v= -0.5cos2x

= -0.5xcosx  -  ∫-0.5cos2x dx

= -0.5xcosx...

Integrate, y= 3x² -1/x

 1{³ 3x² - 1/x dx = [x-lnx]³₁= (3-ln3)-(1-ln1) = 3-ln3-1+0= 2-ln3

f(1)=0

f(x)=(x-1)(x+4)(x-2)