Question

How do you evaluate the integral of `int (x-1)/(x^2+3x+2) dx` from 0 to 1?

Answer 1

`int _0^1 ((x-1)d x)/(x^2+3x+2)=-2,12014923742+C`

Explanation:

`u=x^2+3x+2" "d u=(2x+3)d x`
`x-1=1/2(2x-2+3-3)" "x-1=1/2(2x+3-5)`
`=1/2int (2x+3-5)/(x^2+3x+2)d x" "`
`=1/2(int( (2x+3)d x)/(x^2+3x+2)-int (5d x)/(x^2+3x+2))`
`=1/2(int( d u)/u)-5int (d x)/(x^2+3x+2)`
`1/(x^2+3x+2)=A/(x+2)+B/(x+1)`
`1=A(x+1)+B(x+2)`
`x=-1 " "B=1" "x=-2" "A=-1`
`1/2[l n u-5(-int (d x)/(x+2)+int (d x)/(x+1) )]`
`1/2[l n u-5(l n(x+2)+l n(x+1))]_0^1+C`
`1/2[l n(x^2+3x+2)-5l n(x+2)-5(l n x+1)]_0^1+C`
`1/2[(l n 7-5l n 3-5l n2)-(l n 2-5l n2-5l n1)]+C`
`1/2[(l n7-5l n3-5l n2)-(-4l n2-5l n1)]" "l n1=0`
`1/2[l n7-5l n3-5l n2+4l n2]" "=1/2[l n7-5ln 3-l n2]`
`int _0^1 ((x-1)d x)/(x^2+3x+2)=-2,12014923742+C`

Answer 2

Use partial fractions.

Explanation:

`(x-1)/(x^2+3x+2) = (x-1)/((x+2)(x+1))`

`A/(x+1)+B/(x+2) = (x-1)/((x+2)(x+1))`

`A(x+2)+B(x+1)=x-1`

`Ax+2A+Bx+B = x-1`

`(A+B)x + (2A+B) = 1x+(-1)`

Solve:
`{(A+B=1),(2A+B=-1):}`

`A=-2` and `B=3`

So
`int_0^1 (x-1)/(x^2+3x+2) dx = int_0^1 (-2/(x+1)+3/(x+2) )dx`

`= -2ln(x+1) + 3 ln(x+2)]_0^1`

`= [-2ln(2)+3ln(3)]-[-2ln(1)+3ln(2)]`

`= -5ln(2)+3ln(3)`

`= ln(27/32)`

Rewrite or approximate as you need.