Given:(x²+2x+10)dy=(2x³-x²-4x+1)dx x²+2x+10=(x+1/2)²+(39/4)>0, so we have:
dy/dx=(2x³-x²-4x+1/(x²+2x+10) ··· Eq.1
Let y=∫(dy/dx)dx=∫[(2x³-x²-4x+1)/(x²+2x+10)]dx ··· Eq.2 While,
2x³-x²-4x+1=(2x-5)(x²+2x+10)-7(2x+2)*+65, so Eq.1 can be rewritten:
dy/dx=(2x-5)-7(2x+2)/(x²+2x+10)+65/(x²+2x+10), so Eq.2 also can be rewritten:
y=∫(2x-5)dx-7∫[(2x+2)/(x²+2x+10)]dx+65∫[1/(x²+2x+10)]dx ··· Eq.3 use integration formula for Eq.3
(i). ∫[f'(x)/f(x)]dx=lnlf(x)l, so ∫[(2x+2)/(x²+2x+10)]dx=ln(x²+2x+10)+C1
(ii). In a quadratic equation ax²+bx+c=0, if the discriminant D=b²-4ac is negative(<0),
∫[1/(ax²+bx)]dx=(2/√-D)arctan[(2ax+b)/√-D]+C2. Here, D=2²-4·1·10=-36<0, so
∫[1/(x²+2x+10)]dx=(2/√36)arctan[(2x+2)/√-36]=(1/3)arctan[(x+1)/3] +C3 Thus, Eq.3 can be rewritten:
y=x²-5x-7ln(x²+2x+10)+(65/3)arctan[(x+1)/3]+C ··· Eq.4
CK: Differentiate each term of Eq.4 using derivative rules such as (x^n)'=nx^(n-1), (ln x)'=1/x, (arctanx)'=1/(1+x²) and chainrules.
(x²-5x)'=2x-5, -7[ln(x²+2x+10)]'=-7[(2x+2)/(x²+2x+10)], and 65/3[arctan(x+1)/3]'=65/(x²+2x+10) So, the first derivative Eq.4 is:
y'=dy/dx=(2x-5)-7(2x+2)/(x²+2x+10)+65/(x²+2x+10)=[(2x+5)(x²+2x+10)-7(2x+2)+65}/(x²+2x+10)
=(2x³-x²-4x+1)/(x²+2x+10) (=Eq.1) CKD.
The answer is: y=∫(dy/dx)dx=x²-5x-7ln(x²+2x+10)+(65/3)arctan(x+1)/3+C