How do you integrate #(9x^2 + 1) /( x^2(x − 2)^2)# using partial fractions?

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Answer 1 · 2017-02-28T08:07:42

The answer is #=-1/(4x)-37/(4(x-2))+1/4ln(|x|)-1/4ln(|x-2|)+C#

We perform the decomposition into partial fractions

#(9x^2+1)/(x^2(x-2)^2)=A/(x^2)+B/(x)+C/(x-2)^2+D/(x-2)#

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

As the denominators are the same, we compare the numerators

#9x^2+1=A(x-2)^2+B(x(x-2)^2)+C(x^2)+D(x^2(x-2))#

Let #x=0#, #=>#, #1=4A#, #=>#, #A=1/4#

Let #x=2#, #=>#, #37=4C#, #=>#, #C=37/4#

Coefficients of #x^2#

#9=A-4B+C-2D#

Coeficients of #x#,

#0=-4A+4B#, #=>#, #B=A=1/4#

#1/4-1+37/4-2D=9#

#2D=37/4-3/4-9=34/4-9=-2/4=-1/2#

#D=-1/4#

So,

#(9x^2+1)/(x^2(x-2)^2)=(1/4)/(x^2)+(1/4)/(x)+(37/4)/(x-2)^2+(-1/4)/(x-2)#

Therefore,

#int((9x^2+1)dx)/(x^2(x-2)^2)=1/4intdx/(x^2)+1/4intdx/(x)+37/4intdx/(x-2)^2-1/4intdx/(x-2)#

#=-1/(4x)-37/(4(x-2))+1/4ln(|x|)-1/4ln(|x-2|)+C#