(1) f(x)=3x-2, g(x)=3/x can be interpreted in two ways:
(a) Let h(x)=fg(x)=(3x-2)(3/x)=9-6/x.
Let y=h(x)=9-6/x. So 6/x=9-y, x=6/(9-y).
Let j(y)=x=6/(9-y), so j(x)=h-1(x)=6/(9-x).
Domain of j(x)=h-1(x)=[fg(x)]-1 is all x≠9, that is, {(-∞,9) (9,∞)}.
Range is (-∞,∞).
(b) fg(x) could mean f○g(x)=f(g(x))=f(3/x)=3(3/x)-2=9/x-2.
Let y=h(x)=9/x-2. So 9/x=y+2, x=9/(y+2).
Let j(y)=x=9/(y+2), so j(x)=h-12x)=9/(x+2).
Domain of j(x)=h-1(x)=[fg(x)]-1 is all x≠-2, that is, {(-∞,-2) (-2,∞)}.
Range is (-∞,∞).
(2) 4x3-3x+2 has a higher degree than (2x-1)(x+2)=2x2+3x-2, so we need to divide first:
2x -3 (quotient)
2x2+3x-2 ) 4x3 -3x+2
4x3+6x2-4x
-6x2 +x +2
-6x2 -9x+6
10x-4 (remainder)
CHECK: (2x-3)(2x2+3x-2)+10x-4=4x3+6x2-4x-6x2-9x+6+10x-4=4x3-3x+2✔️
Integrand becomes 2x-3+(10x-4)/((2x-1)(x+2)).
Now we can work out partial fractions:
(10x-4)/((2x-1)(x+2))≡A/(2x-1)+B/(x+2).
Ax+2A+2Bx-B≡10x-4⇒A+2B=10, 2A-B=-4, 4A-2B=-8, 5A=2, A=⅖, B=24/5.
(10x-4)/((2x-1)(x+2))≡⅖(1/(2x-1))+(24/5)(1/(x+2)).
Indefinite integration:
∫(2x-3+(10x-4)/((2x-1)(x+2)))dx=x2-3x+0.2ln|2x-1|+4.8ln|x+2|+C where C is the constant of integration.
If the integration is between limits, the question doesn't display what they are.
If the interval of integration is [a,b] then the result is:
(b-a)(a+b-3)+0.2ln((2b-1)/(2a-1))+4.8ln((b+2)/(a+2)).