Evaluate : `int (x+1)/((x+2)(x+3))*dx`

Explanation

We have `int (x+1)/((x+2)(x+3)).dx`
Let `(x+1)/((x+2)(x+3))=(A)/(x+2)+(B)/(x+3)`
`:.x+1=A(x+3)+B(x+2)`
`:.x+1=(A+B)x+(3A+2B)`
On equating co-efficients of both sides, we get
`A+B=1` ...(1)
`3A+2B=1` ...(2)
Solving equations (1) and (2), we get `A=-1, B=2`
`:.(x+1)/((x+2)(x+3))=(-1)/(x+2)+(2)/(x+3)`
`:.int(x+1)/((x+2)(x+3)).dx=-int(1)/((x+2))dx+2int(1)/((x+3))dx`
`:.int(x+1)/((x+2)(x+3)).dx=-log|x+2|+2log|x+3|+c`