What is the integral of 1/((4-9x^2)^2) ?

The integral of 1/((4-9x^2)^2) is 1/96 ln|3x+2|-1/(48(3x+2))-1/96 ln|3x-2|-1/(48(3x-2))+C

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integral of 1/((4-9x^2)^2)

\int \frac{1}{( 4 - 9 x ^{2} ) ^{2}} d x

\frac{1}{96} ln ∣ 3 x + 2 ∣ - \frac{1}{48 ( 3 x + 2 )} - \frac{1}{96} ln ∣ 3 x - 2 ∣ - \frac{1}{48 ( 3 x - 2 )} + C

Solution steps

\int \frac{1}{( 4 - 9 x ^{2} ) ^{2}} d x

Take the partial fraction of

\frac{1}{( 4 - 9 x ^{2} ) ^{2}} : \frac{1}{32 ( 3 x + 2 )} + \frac{1}{16 ( 3 x + 2 ) ^{2}} - \frac{1}{32 ( 3 x - 2 )} + \frac{1}{16 ( 3 x - 2 ) ^{2}}

= \int \frac{1}{32 ( 3 x + 2 )} + \frac{1}{16 ( 3 x + 2 ) ^{2}} - \frac{1}{32 ( 3 x - 2 )} + \frac{1}{16 ( 3 x - 2 ) ^{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}{32 ( 3 x + 2 )} d x + \int \frac{1}{16 ( 3 x + 2 ) ^{2}} d x - \int \frac{1}{32 ( 3 x - 2 )} d x + \int \frac{1}{16 ( 3 x - 2 ) ^{2}} d x

\int \frac{1}{32 ( 3 x + 2 )} d x = \frac{1}{96} ln ∣ 3 x + 2 ∣

\int \frac{1}{16 ( 3 x + 2 ) ^{2}} d x = - \frac{1}{48 ( 3 x + 2 )}

\int \frac{1}{32 ( 3 x - 2 )} d x = \frac{1}{96} ln ∣ 3 x - 2 ∣

\int \frac{1}{16 ( 3 x - 2 ) ^{2}} d x = - \frac{1}{48 ( 3 x - 2 )}

= \frac{1}{96} ln ∣ 3 x + 2 ∣ - \frac{1}{48 ( 3 x + 2 )} - \frac{1}{96} ln ∣ 3 x - 2 ∣ - \frac{1}{48 ( 3 x - 2 )}

Add a constant to the solution

= \frac{1}{96} ln ∣ 3 x + 2 ∣ - \frac{1}{48 ( 3 x + 2 )} - \frac{1}{96} ln ∣ 3 x - 2 ∣ - \frac{1}{48 ( 3 x - 2 )} + C

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What is the integral of 1/((4-9x^2)^2) ?
The integral of 1/((4-9x^2)^2) is 1/96 ln|3x+2|-1/(48(3x+2))-1/96 ln|3x-2|-1/(48(3x-2))+C