What is the integral of 2/((t^2-4)^2) ?

The integral of 2/((t^2-4)^2) is 2(1/32 ln|t+2|-1/(16(t+2))-1/32 ln|t-2|-1/(16(t-2)))+C

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integral of 2/((t^2-4)^2)

\int \frac{2}{( t ^{2} - 4 ) ^{2}} d t

2 (\frac{1}{32} ln ∣ t + 2 ∣ - \frac{1}{16 ( t + 2 )} - \frac{1}{32} ln ∣ t - 2 ∣ - \frac{1}{16 ( t - 2 )}) + C

Solution steps

\int \frac{2}{( t ^{2} - 4 ) ^{2}} d t

Take the constant out:

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

= 2 \cdot \int \frac{1}{( t ^{2} - 4 ) ^{2}} d t

Take the partial fraction of

\frac{1}{( t ^{2} - 4 ) ^{2}} : \frac{1}{32 ( t + 2 )} + \frac{1}{16 ( t + 2 ) ^{2}} - \frac{1}{32 ( t - 2 )} + \frac{1}{16 ( t - 2 ) ^{2}}

= 2 \cdot \int \frac{1}{32 ( t + 2 )} + \frac{1}{16 ( t + 2 ) ^{2}} - \frac{1}{32 ( t - 2 )} + \frac{1}{16 ( t - 2 ) ^{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

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

\int \frac{1}{32 ( t + 2 )} d t = \frac{1}{32} ln ∣ t + 2 ∣

\int \frac{1}{16 ( t + 2 ) ^{2}} d t = - \frac{1}{16 ( t + 2 )}

\int \frac{1}{32 ( t - 2 )} d t = \frac{1}{32} ln ∣ t - 2 ∣

\int \frac{1}{16 ( t - 2 ) ^{2}} d t = - \frac{1}{16 ( t - 2 )}

= 2 (\frac{1}{32} ln ∣ t + 2 ∣ - \frac{1}{16 ( t + 2 )} - \frac{1}{32} ln ∣ t - 2 ∣ - \frac{1}{16 ( t - 2 )})

Add a constant to the solution

= 2 (\frac{1}{32} ln ∣ t + 2 ∣ - \frac{1}{16 ( t + 2 )} - \frac{1}{32} ln ∣ t - 2 ∣ - \frac{1}{16 ( t - 2 )}) + C

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What is the integral of 2/((t^2-4)^2) ?
The integral of 2/((t^2-4)^2) is 2(1/32 ln|t+2|-1/(16(t+2))-1/32 ln|t-2|-1/(16(t-2)))+C