What is the integral of 3/((t^2-9)^2) ?

The integral of 3/((t^2-9)^2) is 3(1/108 ln|t+3|-1/(36(t+3))-1/108 ln|t-3|-1/(36(t-3)))+C

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integral of 3/((t^2-9)^2)

\int \frac{3}{( t ^{2} - 9 ) ^{2}} d t

3 (\frac{1}{108} ln ∣ t + 3 ∣ - \frac{1}{36 ( t + 3 )} - \frac{1}{108} ln ∣ t - 3 ∣ - \frac{1}{36 ( t - 3 )}) + C

Solution steps

\int \frac{3}{( t ^{2} - 9 ) ^{2}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{( t ^{2} - 9 ) ^{2}} d t

Take the partial fraction of

\frac{1}{( t ^{2} - 9 ) ^{2}} : \frac{1}{108 ( t + 3 )} + \frac{1}{36 ( t + 3 ) ^{2}} - \frac{1}{108 ( t - 3 )} + \frac{1}{36 ( t - 3 ) ^{2}}

= 3 \cdot \int \frac{1}{108 ( t + 3 )} + \frac{1}{36 ( t + 3 ) ^{2}} - \frac{1}{108 ( t - 3 )} + \frac{1}{36 ( t - 3 ) ^{2}} d t

Apply the Sum Rule:

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

= 3 (\int \frac{1}{108 ( t + 3 )} d t + \int \frac{1}{36 ( t + 3 ) ^{2}} d t - \int \frac{1}{108 ( t - 3 )} d t + \int \frac{1}{36 ( t - 3 ) ^{2}} d t)

\int \frac{1}{108 ( t + 3 )} d t = \frac{1}{108} ln ∣ t + 3 ∣

\int \frac{1}{36 ( t + 3 ) ^{2}} d t = - \frac{1}{36 ( t + 3 )}

\int \frac{1}{108 ( t - 3 )} d t = \frac{1}{108} ln ∣ t - 3 ∣

\int \frac{1}{36 ( t - 3 ) ^{2}} d t = - \frac{1}{36 ( t - 3 )}

= 3 (\frac{1}{108} ln ∣ t + 3 ∣ - \frac{1}{36 ( t + 3 )} - \frac{1}{108} ln ∣ t - 3 ∣ - \frac{1}{36 ( t - 3 )})

Add a constant to the solution

= 3 (\frac{1}{108} ln ∣ t + 3 ∣ - \frac{1}{36 ( t + 3 )} - \frac{1}{108} ln ∣ t - 3 ∣ - \frac{1}{36 ( t - 3 )}) + C

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What is the integral of 3/((t^2-9)^2) ?
The integral of 3/((t^2-9)^2) is 3(1/108 ln|t+3|-1/(36(t+3))-1/108 ln|t-3|-1/(36(t-3)))+C