Question

How do you integrate `int (x + 2) / (2x^2 - x - 1)` using partial fractions?

Answer 1

`int (x+2) /(2x^2-x-1) * d x=-1/2 l n(2x+1)+ l n(x-1)+ C`

Explanation:

`int (x+2) /(2x^2-x-1) * d x=?`

`int (x+2)/((2x+1)*(x-1))d x`

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

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

`x=1 rarr" "1+2=A(1-1)+B(2*1+1)`

`3=0+3B" "B=1`

`x=-1/2 rarr" "-1/2+2=A(-1/2-1)+0`

`3/2=A(-3/2)" "A=-1`

`int (x+2) /(2x^2-x-1) * d x=-int (d x)/(2x+1)+int (d x)/(x-1)`

`u=2x+1" "d u=2d x ,d x=(d u)/2" ; "v=x-1" "d v=d x`

`int (x+2) /(2x^2-x-1) * d x=-1/2 int (d u)/u +int (d v)/v`

`int (x+2) /(2x^2-x-1) * d x=-1/2 l n u +l n v +C`

`int (x+2) /(2x^2-x-1) * d x=-1/2 l n(2x+1)+ l n(x-1)+ C`