Whats the intrgral of \int (x^2+x-3)/((x^3+x^2-4x-4)^2)

Question

Whats the intrgral of
`\int (x^2+x-3)/((x^3+x^2-4x-4)^2)`

Answer 1

`\displaystyle\int(x^2+x-3)/((x^3+x^2-4x-4)^2)\,\mathrm dx`

Notice that
`x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)`. Decompose the integrand into partial fractions:


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

`=\frac1{3(x+1)}-(11)/(32(x+2))-\frac1{3(x+1)^2}-\frac1{16(x+2)^2}+\frac1{96(x-2)}+\frac1{48(x-2)^2}`

Integrating term-by-term, you get


`\displaystyle\int(x^2+x-3)/((x^3+x^2-4x-4)^2)\,\mathrm dx`

`=-\frac1{48(x-2)}+\frac1{3(x+1)}+\frac1{16(x+2)}+\frac1{96}\ln|x-2|+\frac13\ln|x+1|-(11)/(32)\ln|x+2|+C`