How do you integrate #int (2x^2+9x-6) / (x^2+x-6)# using partial fractions?

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Answer 1 · 2016-12-18T18:14:36

The answer is #=2x+3ln(∣x+3∣)+4ln(∣x-2∣)+C#

We need to simplify the expression

The denominator is #x^2+x-6=(x+3)(x-2)#

Let's do a long division

#color(white)(aaaa)##2x^2+9x-6##color(white)(aaaa)##∣##x^2+x-6#

#color(white)(aaaa)##2x^2+2x-12##color(white)(aaaa)##∣##2#

#color(white)(aaaaaa)##0+7x+6#

#(2x^2+9x-6)/(x^2+x-6)=2+(7x+6)/(x^2+x-6)#

We can now do the decomposition into partial fractions

#(2x+6)/(x^2+x-6)=(7x+6)/((x+3)(x-2))#

#=A/(x+3)+B/(x-2)#

#=(A(x-2)+B(x+3))/((x+3)(x-2))#

Therefore,

#7x+6=A(x-2)+B(x+3)#

Let #x=2#, #=>#. #20=5B#, #=>#, #B=4#

Let #x=-3#, #=>#, #-15=-5A#, #=>#, #A=3#

#(2x^2+9x-6)/(x^2+x-6)=2+(7x+6)/(x^2+x-6)=2+3/(x+3)+4/(x-2)#

#int((2x^2+9x-6)dx)/(x^2+x-6)=int2dx+3intdx/(x+3)+4intdx/(x-2)#

#=2x+3ln(∣x+3∣)+4ln(∣x-2∣)+C#