Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 15

int_3^4 (x^3-2x^2 - 4)/(x^3 - 2x^2) dx
First we will solve
int (x^3-2x^2 - 4)/(x^3 - 2x^2) dx , then we can apply the limits
=int (x^3-2x^2 - 4)/(x^3 - 2x^2) dx
=int [1- (4/(x^3 - 2x^2)) dx
= int 1 dx - int (4/(x^3 - 2x^2)) dx
= x - 4*int (1/(x^2(x - 2)) dx
1/(x^2(x - 2)) on partial fraction we get
1/((x^2(x - 2)) = ((-1)/4x))+((-1)/(2x^2))+(1/(4(x-2)))
so ,
= x - [4*int (1/(x^2(x - 2)) ] dx
= x-[4*int [((-1)/4x))+((-1)/(2x^2))+(1/(4(x-2)))] dx
= x- [4*int ((-1)/4x))dx+ 4*int((-1)/(2x^2)) dx+ 4*int(1/(4(x-2)))] dx
= x -[4*(-ln(x)/4)+4* (1/(2x)) +4*(ln(x-2)/4)] +c
= x + 4*(ln(x)/4) - 4* (1/(2x)) - 4*(ln(x-2)/4)] +c
so ,
int (x^3-2x^2 - 4)/(x^3 - 2x^2) dx =x + 4*(ln(x)/4) - 4* (1/(2x)) - 4*(ln(x-2)/4)] +c
now,
int_3^4 (x^3-2x^2 - 4)/(x^3 - 2x^2) dx
= [x + 4*(ln(x)/4) - 4* (1/(2x)) - 4*(ln(x-2)/4) ]_3^4
= [4 +(ln(4)) -2* (1/(4)) - (ln(4-2)) ]-[3 +(ln(3))-2* (1/(3)) - (ln(3-2)) ]
=[4 +2*(ln(2))- (1/(2)) - (ln(2)) ]-[3 +(ln(3))-(2/(3)) - (ln(1)) ]
=[4 +(ln(2))- (1/(2)) ]-[3 +(ln(3))-(2/(3)) - (ln(1)) ]
=1+ln(2) - (1/(2))+(2/(3))-ln(3)+0
=(1/2 )+ (2/3)+ln(2)-ln(3)
=(7/6)+ln(2)-ln(3)
= (7/6)+ln(2/3)
is the solution :)