Question

`\int\limits} \, (x^3-x+3)/(x^2+x-2) dx`

Answer 1

First, simplify the quotient:


`\displaystyle\int(x^3-x+3)/(x^2+x-2)\,\mathrm dx=\int\left(x-1+(2x+1)/(x^2+x-2)\right)\,\mathrm dx`

The first two terms are trivial to deal with, while the last term can be taken care of with a substitution of
`y=x^2+x-2`. This gives
`\mathrm dy=(2x+1)\,\mathrm dx`, and you have


`\displaystyle\int (x-1)\,\mathrm dx+\int\frac{\mathrm du}u`

`\frac12x^2-x+\ln|u|+C`

`\frac12x^2-x+\ln|x^2+x-2|+C`