What is the integral of (8-t^3)/((t-3)(t+1)^2) ?

The integral of (8-t^3)/((t-3)(t+1)^2) is -t-19/16 ln|t-3|+3/16 ln|t+1|+9/(4(t+1))+C

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integral of (8-t^3)/((t-3)(t+1)^2)

\int \frac{8 - t ^{3}}{( t - 3 ) ( t + 1 ) ^{2}} d t

- t - \frac{19}{16} ln ∣ t - 3 ∣ + \frac{3}{16} ln ∣ t + 1 ∣ + \frac{9}{4 ( t + 1 )} + C

Solution steps

\int \frac{8 - t ^{3}}{( t - 3 ) ( t + 1 ) ^{2}} d t

Take the partial fraction of

\frac{8 - t ^{3}}{( t - 3 ) ( t + 1 ) ^{2}} : - 1 - \frac{19}{16 ( t - 3 )} + \frac{3}{16 ( t + 1 )} - \frac{9}{4 ( t + 1 ) ^{2}}

= \int - 1 - \frac{19}{16 ( t - 3 )} + \frac{3}{16 ( t + 1 )} - \frac{9}{4 ( t + 1 ) ^{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 1 d t - \int \frac{19}{16 ( t - 3 )} d t + \int \frac{3}{16 ( t + 1 )} d t - \int \frac{9}{4 ( t + 1 ) ^{2}} d t

\int 1 d t = t

\int \frac{19}{16 ( t - 3 )} d t = \frac{19}{16} ln ∣ t - 3 ∣

\int \frac{3}{16 ( t + 1 )} d t = \frac{3}{16} ln ∣ t + 1 ∣

\int \frac{9}{4 ( t + 1 ) ^{2}} d t = - \frac{9}{4 ( t + 1 )}

= - t - \frac{19}{16} ln ∣ t - 3 ∣ + \frac{3}{16} ln ∣ t + 1 ∣ - (- \frac{9}{4 ( t + 1 )})

Simplify

= - t - \frac{19}{16} ln ∣ t - 3 ∣ + \frac{3}{16} ln ∣ t + 1 ∣ + \frac{9}{4 ( t + 1 )}

Add a constant to the solution

= - t - \frac{19}{16} ln ∣ t - 3 ∣ + \frac{3}{16} ln ∣ t + 1 ∣ + \frac{9}{4 ( t + 1 )} + C

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What is the integral of (8-t^3)/((t-3)(t+1)^2) ?
The integral of (8-t^3)/((t-3)(t+1)^2) is -t-19/16 ln|t-3|+3/16 ln|t+1|+9/(4(t+1))+C