Question

How do you integrate `int 5/ ( (2x-3) (x+1) ) dx` using partial fractions?

Answer 1

`int 5/((2x-3)(x+1))d x=l n(2x-3)-l n(x+1) +C`

Explanation:

`int 5/((2x-3)(x+1))d x=5int(d x)/((2x-3)(x+1))`
`1/((2x-3)(x+1))=A/(2x-3)+B/(x+1)`
`1/((2x-3)(x+1))=(A(x+1)+B(2x-3))/((2x-3)(x+1))`
`"for x="-1 =>" " 1=B(2*(-1)-3)" " B=-1/5`
`"for x="3/2 =>" " 1=A(3/2+1)" "A=2/5`
`5int(d x)/((2x-3)(x+1))=5(int(2d x)/(5(2x-3))-int(d x)/(5(x+1)))`
`=int (2d x)/((2x-3))-int (d x)/((x+1))`
`u=2x-3" "d u=2d x" "v=x+1" "d v=d x`
`=int (d u)/u-int (d v)/v+C`
`int 5/((2x-3)(x+1))d x=l n u-l n v+C`
`int 5/((2x-3)(x+1))d x=l n(2x-3)-l n(x+1) +C`