How do you integrate $\frac{9 x^{2} + 1}{x^{2} {(x - 2)}^{2}}$ $(9x^2 + 1) /( x^2(x − 2)^2)$ using partial fractions?

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How do you integrate $\frac{9 x^{2} + 1}{x^{2} {(x - 2)}^{2}}$ $(9x^2 + 1) /( x^2(x − 2)^2)$ using partial fractions?

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Answer 1 · 2017-02-28T08:07:42

The answer is $= - \frac{1}{4 x} - \frac{37}{4 (x - 2)} + \frac{1}{4} ln (| x |) - \frac{1}{4} ln (| x - 2 |) + C$ $=-1/(4x)-37/(4(x-2))+1/4ln(|x|)-1/4ln(|x-2|)+C$

Explanation:

We perform the decomposition into partial fractions

$\frac{9 x^{2} + 1}{x^{2} {(x - 2)}^{2}} = \frac{A}{x^{2}} + \frac{B}{x} + \frac{C}{{(x - 2)}^{2}} + \frac{D}{x - 2}$ $(9x^2+1)/(x^2(x-2)^2)=A/(x^2)+B/(x)+C/(x-2)^2+D/(x-2)$

$= \frac{A {(x - 2)}^{2} + B (x {(x - 2)}^{2}) + C (x^{2}) + D (x^{2} (x - 2))}{x^{2} {(x - 2)}^{2}}$ $=(A(x-2)^2+B(x(x-2)^2)+C(x^2)+D(x^2(x-2)))/(x^2(x-2)^2)$

As the denominators are the same, we compare the numerators

$9 x^{2} + 1 = A {(x - 2)}^{2} + B (x {(x - 2)}^{2}) + C (x^{2}) + D (x^{2} (x - 2))$ $9x^2+1=A(x-2)^2+B(x(x-2)^2)+C(x^2)+D(x^2(x-2))$

Let $x = 0$ $x=0$ , $\Rightarrow$ $=>$ , $1 = 4 A$ $1=4A$ , $\Rightarrow$ $=>$ , $A = \frac{1}{4}$ $A=1/4$

Let $x = 2$ $x=2$ , $\Rightarrow$ $=>$ , $37 = 4 C$ $37=4C$ , $\Rightarrow$ $=>$ , $C = \frac{37}{4}$ $C=37/4$

Coefficients of $x^{2}$ $x^2$

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

Coeficients of $x$ $x$ ,

$0 = - 4 A + 4 B$ $0=-4A+4B$ , $\Rightarrow$ $=>$ , $B = A = \frac{1}{4}$ $B=A=1/4$

$\frac{1}{4} - 1 + \frac{37}{4} - 2 D = 9$ $1/4-1+37/4-2D=9$

$2 D = \frac{37}{4} - \frac{3}{4} - 9 = \frac{34}{4} - 9 = - \frac{2}{4} = - \frac{1}{2}$ $2D=37/4-3/4-9=34/4-9=-2/4=-1/2$

$D = - \frac{1}{4}$ $D=-1/4$

So,

$\frac{9 x^{2} + 1}{x^{2} {(x - 2)}^{2}} = \frac{\frac{1}{4}}{x^{2}} + \frac{\frac{1}{4}}{x} + \frac{\frac{37}{4}}{{(x - 2)}^{2}} + \frac{- \frac{1}{4}}{x - 2}$ $(9x^2+1)/(x^2(x-2)^2)=(1/4)/(x^2)+(1/4)/(x)+(37/4)/(x-2)^2+(-1/4)/(x-2)$

Therefore,

$\int \frac{(9 x^{2} + 1) d x}{x^{2} {(x - 2)}^{2}} = \frac{1}{4} \int \frac{d x}{x^{2}} + \frac{1}{4} \int \frac{d x}{x} + \frac{37}{4} \int \frac{d x}{{(x - 2)}^{2}} - \frac{1}{4} \int \frac{d x}{x - 2}$ $int((9x^2+1)dx)/(x^2(x-2)^2)=1/4intdx/(x^2)+1/4intdx/(x)+37/4intdx/(x-2)^2-1/4intdx/(x-2)$

$= - \frac{1}{4 x} - \frac{37}{4 (x - 2)} + \frac{1}{4} ln (| x |) - \frac{1}{4} ln (| x - 2 |) + C$ $=-1/(4x)-37/(4(x-2))+1/4ln(|x|)-1/4ln(|x-2|)+C$

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