What is the integral of (2t^3-4t^2+18t)/(t^2+9) ?

The integral of (2t^3-4t^2+18t)/(t^2+9) is t^2-4t+12arctan(t/3)+9+C

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integral of (2t^3-4t^2+18t)/(t^2+9)

\int \frac{2 t ^{3} - 4 t ^{2} + 18 t}{t ^{2} + 9} d t

t^{2} - 4 t + 12 arctan (\frac{t}{3}) + 9 + C

Solution steps

\int \frac{2 t ^{3} - 4 t ^{2} + 18 t}{t ^{2} + 9} d t

Expand

\frac{2 t ^{3} - 4 t ^{2} + 18 t}{t ^{2} + 9} : \frac{2 t ^{3}}{t ^{2} + 9} - \frac{4 t ^{2}}{t ^{2} + 9} + \frac{18 t}{t ^{2} + 9}

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{2 t ^{3}}{t ^{2} + 9} d t - \int \frac{4 t ^{2}}{t ^{2} + 9} d t + \int \frac{18 t}{t ^{2} + 9} d t

\int \frac{2 t ^{3}}{t ^{2} + 9} d t = t^{2} + 9 - 9 ln t^{2} + 9

\int \frac{4 t ^{2}}{t ^{2} + 9} d t = 12 (- arctan (\frac{t}{3}) + \frac{t}{3})

\int \frac{18 t}{t ^{2} + 9} d t = 9 ln t^{2} + 9

= t^{2} + 9 - 9 ln t^{2} + 9 - 12 (- arctan (\frac{t}{3}) + \frac{t}{3}) + 9 ln t^{2} + 9

Simplify

t^{2} + 9 - 9 ln t^{2} + 9 - 12 (- arctan (\frac{t}{3}) + \frac{t}{3}) + 9 ln t^{2} + 9 : t^{2} - 4 t + 12 arctan (\frac{t}{3}) + 9

= t^{2} - 4 t + 12 arctan (\frac{t}{3}) + 9

Add a constant to the solution

= t^{2} - 4 t + 12 arctan (\frac{t}{3}) + 9 + C

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What is the integral of (2t^3-4t^2+18t)/(t^2+9) ?
The integral of (2t^3-4t^2+18t)/(t^2+9) is t^2-4t+12arctan(t/3)+9+C