Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 10

inty/((y+4)(2y-1)) dy
sol:-
inty/((y+4)(2y-1)) dy
First take partial fractions of y/((y+4)(2y-1))
y/((y+4)(2y-1)) = (A/(y+4)) +(B/(2y-1))
= ((A*(2y-1))+(B*(y+4)))/((y+4)(2y-1))
Now equate the numerators we get
y=((A*(2y-1))+(B*(y+4)))
= 2Ay-A+By+4B
= (2A+B)y + (4B-A)
Now equating the co efficients of y and the constants we get
2A+B=1 , 4B-A=0
=> 4B=A
2A+B=1
=> 2(4B) +B=1
=> 8B+B=1
B=1/9
so A= 4/9
Then
y/((y+4)(2y-1)) = (A/(y+4)) +(B/(2y-1))
= (4/(9(y+4))) +(1/(9(2y-1)))
Now,
inty/((y+4)(2y-1)) dy
= int [(4/(9(y+4))) +(1/(9(2y-1)))]dy
= int (4/(9(y+4))) dy + int (1/(9(2y-1)))dy
= (4/9) ln(y+4) + (1/9) *(1/2)(ln(2y-1)) +c
= (4/9) ln(y+4) + (1/18)(ln(2y-1)) +c
is the solution
:)