How do you integrate #int ( x^2 + 7x +3) /( x^2 (x + 3))# using partial fractions?

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Answer 1 · 2017-01-19T13:37:33

The answer is #=-1/x+2ln(∣x∣)-ln(∣x+3∣)+C#

Let's perform the decomposition into partial fractions

#(x^2+7x+3)/(x^2(x+3))=A/x^2+B/x+C/(x+3)#

#=(A(x+3)+Bx(x+3)+Cx^2)/(x^2(x+3))#

Equalising the denominators

#x^2+7x+3=A(x+3)+Bx(x+3)+Cx^2#

Let #x=0#, #=>#, #3=3A#, #=>#, #A=1#

Let #x=-3#, #=>#, #-9=9C#, #=>#, #C=-1#

Coefficients of #x^2#

#1=B+C#, #=>#, #B=1-C=2#

Therefore,

#(x^2+7x+3)/(x^2(x+3))=1/x^2+2/x-1/(x+3)#

So,

#int((x^2+7x+3)dx)/(x^2(x+3))=intdx/x^2+2intdx/x-intdx/(x+3)#

#=-1/x+2ln(∣x∣)-ln(∣x+3∣)+C#