How do you integrate #\int \frac { 5x - 3} { x ^ { 2} + 5x + 6} dx#?

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Answer 1 · 2018-05-06T13:44:05

The answer is #=-13ln(|x+2|)+18ln(|x+3|)+C#

Factorise denominator is

#x^2+5x+6=(x+2)(x+3)#

The degree of the numerator is #># than the degree of the denominaor

Perform the decomposition into partial fractions

#(5x-3)/(x^2+5x+6)=(5x-3)/((x+2)(x+3))#

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

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

The denominators are the same, compare the numerators

#5x-3=A(x+3)+B(x+2)#

Let #x=-2#, #=>#, #-13=A#

Let #x=-3#, #=>#, #-18=-B#, #=>#, #B=18#

Therefore,

#(5x-3)/(x^2+5x+6)=-13/(x+2)+18/(x+3)#

So,

#int((5x-3)dx)/(x^2+5x+6)=int(-13dx)/(x+2)+int(18dx)/(x+3)#

#=-13ln(|x+2|)+18ln(|x+3|)+C#