\(\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx\)
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:
\(\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}\)
\(=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}\)
Integrating term-by-term, you get
\(\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx\)
\(=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C\)
