What is the integral of (4t^3-t^2+36t)/(t^2+9) ?

The integral of (4t^3-t^2+36t)/(t^2+9) is 2t^2-t+3arctan(t/3)+18+C

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integral of (4t^3-t^2+36t)/(t^2+9)

\int \frac{4 t ^{3} - t ^{2} + 36 t}{t ^{2} + 9} d t

2 t^{2} - t + 3 arctan (\frac{t}{3}) + 18 + C

Solution steps

\int \frac{4 t ^{3} - t ^{2} + 36 t}{t ^{2} + 9} d t

Expand

\frac{4 t ^{3} - t ^{2} + 36 t}{t ^{2} + 9} : \frac{4 t ^{3}}{t ^{2} + 9} - \frac{t ^{2}}{t ^{2} + 9} + \frac{36 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{4 t ^{3}}{t ^{2} + 9} d t - \int \frac{t ^{2}}{t ^{2} + 9} d t + \int \frac{36 t}{t ^{2} + 9} d t

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

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

\int \frac{36 t}{t ^{2} + 9} d t = 18 ln t^{2} + 9

= 2 (t^{2} + 9 - 9 ln t^{2} + 9) - (- 3 arctan (\frac{t}{3}) + t) + 18 ln t^{2} + 9

Simplify

2 (t^{2} + 9 - 9 ln t^{2} + 9) - (- 3 arctan (\frac{t}{3}) + t) + 18 ln t^{2} + 9 : 2 t^{2} - t + 3 arctan (\frac{t}{3}) + 18

= 2 t^{2} - t + 3 arctan (\frac{t}{3}) + 18

Add a constant to the solution

= 2 t^{2} - t + 3 arctan (\frac{t}{3}) + 18 + C

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What is the integral of (4t^3-t^2+36t)/(t^2+9) ?
The integral of (4t^3-t^2+36t)/(t^2+9) is 2t^2-t+3arctan(t/3)+18+C