What is the integral of 5/((t^2-25)^2) ?

The integral of 5/((t^2-25)^2) is 5(1/500 ln|t+5|-1/(100(t+5))-1/500 ln|t-5|-1/(100(t-5)))+C

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integral of 5/((t^2-25)^2)

\int \frac{5}{( t ^{2} - 25 ) ^{2}} d t

5 (\frac{1}{500} ln ∣ t + 5 ∣ - \frac{1}{100 ( t + 5 )} - \frac{1}{500} ln ∣ t - 5 ∣ - \frac{1}{100 ( t - 5 )}) + C

Solution steps

\int \frac{5}{( t ^{2} - 25 ) ^{2}} d t

Take the constant out:

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

= 5 \cdot \int \frac{1}{( t ^{2} - 25 ) ^{2}} d t

Take the partial fraction of

\frac{1}{( t ^{2} - 25 ) ^{2}} : \frac{1}{500 ( t + 5 )} + \frac{1}{100 ( t + 5 ) ^{2}} - \frac{1}{500 ( t - 5 )} + \frac{1}{100 ( t - 5 ) ^{2}}

= 5 \cdot \int \frac{1}{500 ( t + 5 )} + \frac{1}{100 ( t + 5 ) ^{2}} - \frac{1}{500 ( t - 5 )} + \frac{1}{100 ( t - 5 ) ^{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

= 5 (\int \frac{1}{500 ( t + 5 )} d t + \int \frac{1}{100 ( t + 5 ) ^{2}} d t - \int \frac{1}{500 ( t - 5 )} d t + \int \frac{1}{100 ( t - 5 ) ^{2}} d t)

\int \frac{1}{500 ( t + 5 )} d t = \frac{1}{500} ln ∣ t + 5 ∣

\int \frac{1}{100 ( t + 5 ) ^{2}} d t = - \frac{1}{100 ( t + 5 )}

\int \frac{1}{500 ( t - 5 )} d t = \frac{1}{500} ln ∣ t - 5 ∣

\int \frac{1}{100 ( t - 5 ) ^{2}} d t = - \frac{1}{100 ( t - 5 )}

= 5 (\frac{1}{500} ln ∣ t + 5 ∣ - \frac{1}{100 ( t + 5 )} - \frac{1}{500} ln ∣ t - 5 ∣ - \frac{1}{100 ( t - 5 )})

Add a constant to the solution

= 5 (\frac{1}{500} ln ∣ t + 5 ∣ - \frac{1}{100 ( t + 5 )} - \frac{1}{500} ln ∣ t - 5 ∣ - \frac{1}{100 ( t - 5 )}) + C

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What is the integral of 5/((t^2-25)^2) ?
The integral of 5/((t^2-25)^2) is 5(1/500 ln|t+5|-1/(100(t+5))-1/500 ln|t-5|-1/(100(t-5)))+C