What is the integral of (3-7t)/((t^2+12t+40)^2) ?

The integral of (3-7t)/((t^2+12t+40)^2) is

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integral of (3-7t)/((t^2+12t+40)^2)

\int \frac{3 - 7 t}{( t ^{2} + 12 t + 40 ) ^{2}} d t

\frac{3}{16} (arctan (\frac{1}{2} (t + 6)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (t + 6)))) - 7 (\frac{1}{16} t (arctan (\frac{1}{2} (t + 6)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (t + 6)))) - \frac{1}{16} arctan (\frac{t + 6}{2}) (t + 6)) + C

Solution steps

\int \frac{3 - 7 t}{( t ^{2} + 12 t + 40 ) ^{2}} d t

Expand

\frac{3 - 7 t}{( t ^{2} + 12 t + 40 ) ^{2}} : \frac{3}{( t ^{2} + 12 t + 40 ) ^{2}} - \frac{7 t}{( t ^{2} + 12 t + 40 ) ^{2}}

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{3}{( t ^{2} + 12 t + 40 ) ^{2}} d t - \int \frac{7 t}{( t ^{2} + 12 t + 40 ) ^{2}} d t

\int \frac{3}{( t ^{2} + 12 t + 40 ) ^{2}} d t = \frac{3}{16} (arctan (\frac{1}{2} (t + 6)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (t + 6))))

\int \frac{7 t}{( t ^{2} + 12 t + 40 ) ^{2}} d t = 7 (\frac{1}{16} t (arctan (\frac{1}{2} (t + 6)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (t + 6)))) - \frac{1}{16} arctan (\frac{t + 6}{2}) (t + 6))

= \frac{3}{16} (arctan (\frac{1}{2} (t + 6)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (t + 6)))) - 7 (\frac{1}{16} t (arctan (\frac{1}{2} (t + 6)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (t + 6)))) - \frac{1}{16} arctan (\frac{t + 6}{2}) (t + 6))

Add a constant to the solution

= \frac{3}{16} (arctan (\frac{1}{2} (t + 6)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (t + 6)))) - 7 (\frac{1}{16} t (arctan (\frac{1}{2} (t + 6)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (t + 6)))) - \frac{1}{16} arctan (\frac{t + 6}{2}) (t + 6)) + C

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What is the integral of (3-7t)/((t^2+12t+40)^2) ?
The integral of (3-7t)/((t^2+12t+40)^2) is