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,

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

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

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