Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 16

int_0^1 (x^3 - 4x - 10)/(x^2 - x - 6) dx
First simplify
(x^3 - 4x - 10)/(x^2 - x - 6)
By dividing practically we get the quotient as (x+1) and remainder (3x-4) ,so we can write it as
(x^3 - 4x - 10)/(x^2 - x - 6)
= (x+1)+ ((3x-4)/(x^2 - x - 6))
so now we can write
int_0^1 (x^3 - 4x - 10)/(x^2 - x - 6) dx
=int_0^1 (x+1)+ ((3x-4)/(x^2 - x - 6))dx
Now the simplification is as follows :
First just integrate and then apply the limits
so,
int (x+1)+ ((3x-4)/(x^2 - x - 6))dx
=int (x+1) dx+ int ((3x-4)/(x^2 - x - 6))dx
= ((x^2)/2) +x +int ((3x-4)/(x^2 - x - 6))dx
= ((x^2)/2) +x +int ((3x-4)/((x+2) ( x - 3)))dx
Using partial fractions we get
(3x-4)/((x+2) ( x - 3))= A/(x+2) +b/(x-3)
=> On solving we get A=2 B= 1 so,
(3x-4)/((x+2) ( x - 3))= 2/(x+2) +1/(x-3)
so,
=>((x^2)/2) +x +int ((3x-4)/((x+2) ( x - 3)))dx
=((x^2)/2) +x +int ((2/(x+2)) +(1/(x-3)))dx
= ((x^2)/2) +x +int (2/(x+2)) dx + int (1/(x-3))dx
=((x^2)/2) +x + 2*ln(x+2) + ln(x-3)
Now apply the limits o to 1 we get
=[((x^2)/2) +x + 2*ln(x+2) + ln(x-3)]_0^1
=[((1^2)/2) +1 + 2*ln(1+2) + ln(1-3)] -[((0^2)/2) + 0 + 2*ln(0+2) + ln(0-3)]
= (1/2) + 1 + 2*ln(3) + ln(-2) -[2*ln(2) +ln(-3)]
= 3/2 + 2*[ln(3) - ln(2)] + ln(-2) -ln(-3)
= 3/2 + 2*[ln(3/2)] + ln(-2) -ln(-3)
= 3/2 + 2*[ln(3/2)] + ln(-2/-3)
= 3/2 + 2*[ln(3/2)] + ln(2/3)
= 3/2 + 2*[ln(3/2)] - ln(1/(2/3))
= 3/2 + 2*[ln(3/2)] - ln(3/2)
= 3/2 +ln(3/2)
so ,
int_0^1 (x^3 - 4x - 10)/(x^2 - x - 6) dx =3/2 +ln(3/2)

:)