How to integrate [(x + C) / (AX ^ 2-bx)] DX?

How to integrate [(x + C) / (AX ^ 2-bx)] DX?

x+c=x-(b/a)+(c+b/a)
Integrand function to 1 / ax + (c + B / a) / (AX ^ 2-bx)
The first term integral is alnx, and the second term is column term
1/(ax^2-bx)=1/(x(ax-b))=a/b(1/(ax-b)-1/ax)
It should be next

Integral calculation ∫ √ (AX BX & # 178;) DX

∫ √(ax-bx²) dx
= ∫ √{-[√b*x - a/(2√b)]² + a²/(4b)} dx
Let u = √b*x - a/(2√b) and du = √b dx
==> (1/√b)∫ √[-u² + a²/(4b)] du
Let u = a*siny/(2√b) and du = a*cosy/(2√b) dy,√[-u² + a²/(4b)] = a*cosy/(2√b)
==> a²/[4b^(3*2)] * ∫ cos²y dy
= a²/[4b^(3/2)] * (1/2)∫ (1 + cos2y) dy
= a²/[4b^(3/2)] * (1/2)(y + 1/2 * sin2y) + C
= a²/[8b^(3/2)] * y + a²/[8b^(3/2)] * sinycosy + C
= a²/[8b^(3/2)] * arcsin(2√b*u / a) + a²/[8b^(3/2)] * (2√b*u / a) * √(a²-4bu²) / a
= -a/[8b^(3/2)] * [(2/a)(a-2bx)√(abx-b²x²) + a*arcsin(1-2bx/a)] + C

Let ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e = (X-2) ^ 4
Find the value of algebraic formula (1) a + B + C + D + E; (2) a + C

The latter is expanded to x ^ 4-8x ^ 3 + 12x ^ 2-12x + 16
So a = 1
b=-8
c=12
d=-12
e=16
a+b+c+d+e=25
a+c=13