What is the integral of (t^2-2)/((t^2-t-6)^2) ?

The integral of (t^2-2)/((t^2-t-6)^2) is -16/125 ln|t+2|-2/(25(t+2))+16/125 ln|t-3|-7/(25(t-3))+C

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integral of (t^2-2)/((t^2-t-6)^2)

\int \frac{t ^{2} - 2}{( t ^{2} - t - 6 ) ^{2}} d t

- \frac{16}{125} ln ∣ t + 2 ∣ - \frac{2}{25 ( t + 2 )} + \frac{16}{125} ln ∣ t - 3 ∣ - \frac{7}{25 ( t - 3 )} + C

Solution steps

\int \frac{t ^{2} - 2}{( t ^{2} - t - 6 ) ^{2}} d t

Take the partial fraction of

\frac{t ^{2} - 2}{( t ^{2} - t - 6 ) ^{2}} : - \frac{16}{125 ( t + 2 )} + \frac{2}{25 ( t + 2 ) ^{2}} + \frac{16}{125 ( t - 3 )} + \frac{7}{25 ( t - 3 ) ^{2}}

= \int - \frac{16}{125 ( t + 2 )} + \frac{2}{25 ( t + 2 ) ^{2}} + \frac{16}{125 ( t - 3 )} + \frac{7}{25 ( 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

= - \int \frac{16}{125 ( t + 2 )} d t + \int \frac{2}{25 ( t + 2 ) ^{2}} d t + \int \frac{16}{125 ( t - 3 )} d t + \int \frac{7}{25 ( t - 3 ) ^{2}} d t

\int \frac{16}{125 ( t + 2 )} d t = \frac{16}{125} ln ∣ t + 2 ∣

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

\int \frac{16}{125 ( t - 3 )} d t = \frac{16}{125} ln ∣ t - 3 ∣

\int \frac{7}{25 ( t - 3 ) ^{2}} d t = - \frac{7}{25 ( t - 3 )}

= - \frac{16}{125} ln ∣ t + 2 ∣ - \frac{2}{25 ( t + 2 )} + \frac{16}{125} ln ∣ t - 3 ∣ - \frac{7}{25 ( t - 3 )}

Add a constant to the solution

= - \frac{16}{125} ln ∣ t + 2 ∣ - \frac{2}{25 ( t + 2 )} + \frac{16}{125} ln ∣ t - 3 ∣ - \frac{7}{25 ( t - 3 )} + C

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What is the integral of (t^2-2)/((t^2-t-6)^2) ?
The integral of (t^2-2)/((t^2-t-6)^2) is -16/125 ln|t+2|-2/(25(t+2))+16/125 ln|t-3|-7/(25(t-3))+C