Question

Whats the intrgral of D%20" id="TexFormula1" title=" \int \frac{x^2+x-3}{(x^3+x^2-4x-4)^2} " alt=" \int \frac{x^2+x-3}{(x^3+x^2-4x-4)^2} " align="absmiddle" class="latex-formula">

Answer 1

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

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

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