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

User Stam
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1 Answer

4 votes

\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
User Flatliner DOA
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