What is the integral of (x^2)/((x^2-9)^2) ?

The integral of (x^2)/((x^2-9)^2) is -1/12 ln|x+3|-1/(4(x+3))+1/12 ln|x-3|-1/(4(x-3))+C

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integral of (x^2)/((x^2-9)^2)

\int \frac{x ^{2}}{( x ^{2} - 9 ) ^{2}} d x

- \frac{1}{12} ln ∣ x + 3 ∣ - \frac{1}{4 ( x + 3 )} + \frac{1}{12} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )} + C

Solution steps

\int \frac{x ^{2}}{( x ^{2} - 9 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x ^{2} - 9 ) ^{2}} : - \frac{1}{12 ( x + 3 )} + \frac{1}{4 ( x + 3 ) ^{2}} + \frac{1}{12 ( x - 3 )} + \frac{1}{4 ( x - 3 ) ^{2}}

= \int - \frac{1}{12 ( x + 3 )} + \frac{1}{4 ( x + 3 ) ^{2}} + \frac{1}{12 ( x - 3 )} + \frac{1}{4 ( x - 3 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{12 ( x + 3 )} d x + \int \frac{1}{4 ( x + 3 ) ^{2}} d x + \int \frac{1}{12 ( x - 3 )} d x + \int \frac{1}{4 ( x - 3 ) ^{2}} d x

\int \frac{1}{12 ( x + 3 )} d x = \frac{1}{12} ln ∣ x + 3 ∣

\int \frac{1}{4 ( x + 3 ) ^{2}} d x = - \frac{1}{4 ( x + 3 )}

\int \frac{1}{12 ( x - 3 )} d x = \frac{1}{12} ln ∣ x - 3 ∣

\int \frac{1}{4 ( x - 3 ) ^{2}} d x = - \frac{1}{4 ( x - 3 )}

= - \frac{1}{12} ln ∣ x + 3 ∣ - \frac{1}{4 ( x + 3 )} + \frac{1}{12} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )}

Add a constant to the solution

= - \frac{1}{12} ln ∣ x + 3 ∣ - \frac{1}{4 ( x + 3 )} + \frac{1}{12} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )} + C

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What is the integral of (x^2)/((x^2-9)^2) ?
The integral of (x^2)/((x^2-9)^2) is -1/12 ln|x+3|-1/(4(x+3))+1/12 ln|x-3|-1/(4(x-3))+C