Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 20

int(x^2-5x+16)/((2x+1)(x-2)^2)dx2
Let's first express the integrand as sum of proper rational expressions by applying partial fraction decomposition,
(x^2-5x+16)/((2x+1)(x-2)^2)=A/(2x+1)+B/(x-2)+C/(x-2)^2
=(A(x-2)^2+B(2x+1)(x-2)+C(2x+1))/((2x+1)(x-2)^2)
=(A(x^2-4x+4)+B(2x^2-4x+x-2)+C(2x+1))/((2x+1)(x-2)^2)
=(A(x^2-4x+4)+B(2x^2-3x-2)+C(2x+1))/((2x+1)(x-2)^2)
=(x^2(A+2B)+x(-4A-3B+2C)+4A-2B+C)/((2x+1)(x-2)^2)
Now equate the coefficients of the polynomial in the numerator on the both sides,
A+2B=1 --------------------------------(1)
-4A-3B+2C=-5 -----------------(2)
4A-2B+C=16 ----------------------(3)
Now let's solve the above three equations by the method of substitution,
From equation 1 :A=1-2B
Substitute the above value of A in equation 2 ,
-4(1-2B)-3B+2C=-5
-4+8B-3B+2C=-5
5B+2C=-5+4
5B+2C=-1 ------------------------ (4)
Now substitute the value of A in equation 3,
4(1-2B)-2B+C=16
4-8B-2B+C=16
4-10B+C=16
-10B+C=16-4
-10B+C=12 -----------------------(5)
Now solve the equations 4 and 5 by the method of elimination,
Multiply equation 4 by 2,
10B+4C=-2 ----------------------(6)
Now add the equations 5 and 6,
5C=12-2=10
C=10/5=2
Plug the value of C in equation 5.
-10B+2=12
-10B=12-2=10
B=10/-10=-1
Plug the value of B in equation 1.
A+2(-1)=1
A-2=1
A=1+2=3
:.int(x^2-5x+6)/((2x+1)(x-2)^2)dx=int(3/(2x+1)+(-1)/(x-2)+2/(x-2)^2)dx
=3int1/(2x+1)dx-int1/(x-2)dx+2int1/(x-2)^2dx
Now let's evaluate the above three integrals,
int1/(2x+1)dx
Let's apply the integral substitution:u=2x+1
du=2dx
=int1/(2u)du
=1/2ln|u|
Substitute back u=2x+1,
=1/2ln|2x+1|
Now let's evaluate int1/(x-2)dx
apply integral substitution: v=x-2
dv=dx
=int1/vdv
=ln|v|
substitute back v=x-2,
=ln|x-2|
Now let's evaluate integral int1/(x-2)^2dx
apply the integral substitution: t=x-2
dt=dx
int1/t^2dt
=intt^-2dt
=t^(-2+1)/(-2+1)
=-1/t
Substitute back t=x-2,
=-1/(x-2)
:.int(x^2-5x+16)/((2x+1)(x-2)^2)dx=3(1/2ln|2x+1|)-1ln|x-2|+2(-1/(x-2))
=3/2ln|2x+1|-ln|x-2|-2/(x-2)+C where C is a constant