`int_0^1 (x^3 - 4x - 10)/(x^2 - x - 6)` Evaluate the integral

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Chapter 7, 7.4 - Problem 16 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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`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)`

:)                  

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