How do you evaluate the integral $int (2x^2+x-5)/((x-3)(x+2))$ ?

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Answer 1 · 2018-02-19T12:21:14

$int (2x^2+x-5)/(x^2-x-6)*dx=2x+2ln(x-3)-ln(x+2)+C$

$int (2x^2+x-5)/(x^2-x-6)*dx$

= $int 2dx$ + $int (x+7)/(x^2-x-6)*dx$

= $2x+int (x+7)/((x-3)*(x+2))*dx$

= $2x+int (2x+4)/((x-3)(x+2))*dx$ - $int (x-3)/((x-3)(x+2))*dx$

= $2x+int (2dx)/(x-3)$ - $int (dx)/(x+2)$

= $2x+2ln(x-3)-ln(x+2)+C$

How do you evaluate the integral int (2x^2+x-5)/((x-3)(x+2))int (2x^2+x-5)/((x-3)(x+2))?