How do you integrate #(x+7)/(x^2(x+2))# using partial fractions?

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Answer 1 · 2017-02-08T18:04:26

The answer is #=-7/(2x)-5/4ln(|x|)+5/4ln(|x+2|)+C#

Let's perform the decomposition into partial fractions

#(x+7)/(x^2(x+2))=A/(x^2)+B/(x)+C/(x+2)#

#=(A(x+2)+B(x(x+2))+C(x^2))/(x^2(x+2))#

As the denominators are the same, we compare the numerators

#x+7=A(x+2)+B(x(x+2))+C(x^2)#

Let #A=0#, #=>#, #7=2A#, #=>#, #A=7/2#

Let #x=-2#, #=>#, #5=4C#, #=>#, #C=5/4#

Coefficients of #x#

#1=A+2B#, #=>#, #2B=1-A=1-7/2=-5/2#

#B=-5/4#

Therefore,

#(x+7)/(x^2(x+2))=(7/2)/(x^2)+(-5/4)/(x)+(5/4)/(x+2)#

So,

#int((x+7)dx)/(x^2(x+2))=7/2intdx/(x^2)-5/4intdx/(x)+5/4intdx/(x+2)#

#=-7/(2x)-5/4ln(|x|)+5/4ln(|x+2|)+C#