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

User Sukotto
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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
User Tassadaque
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