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

1 Answer

Feb 19, 2018

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

Explanation:

#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#

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