What is the integral of (6t^3-4t^2+24t)/(t^2+4) ?

The integral of (6t^3-4t^2+24t)/(t^2+4) is 3t^2-4t+8arctan(t/2)+12+C

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integral of (6t^3-4t^2+24t)/(t^2+4)

\int \frac{6 t ^{3} - 4 t ^{2} + 24 t}{t ^{2} + 4} d t

3 t^{2} - 4 t + 8 arctan (\frac{t}{2}) + 12 + C

Solution steps

\int \frac{6 t ^{3} - 4 t ^{2} + 24 t}{t ^{2} + 4} d t

Expand

\frac{6 t ^{3} - 4 t ^{2} + 24 t}{t ^{2} + 4} : \frac{6 t ^{3}}{t ^{2} + 4} - \frac{4 t ^{2}}{t ^{2} + 4} + \frac{24 t}{t ^{2} + 4}

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{6 t ^{3}}{t ^{2} + 4} d t - \int \frac{4 t ^{2}}{t ^{2} + 4} d t + \int \frac{24 t}{t ^{2} + 4} d t

\int \frac{6 t ^{3}}{t ^{2} + 4} d t = 3 (t^{2} + 4 - 4 ln t^{2} + 4)

\int \frac{4 t ^{2}}{t ^{2} + 4} d t = 8 (- arctan (\frac{t}{2}) + \frac{t}{2})

\int \frac{24 t}{t ^{2} + 4} d t = 12 ln t^{2} + 4

= 3 (t^{2} + 4 - 4 ln t^{2} + 4) - 8 (- arctan (\frac{t}{2}) + \frac{t}{2}) + 12 ln t^{2} + 4

Simplify

3 (t^{2} + 4 - 4 ln t^{2} + 4) - 8 (- arctan (\frac{t}{2}) + \frac{t}{2}) + 12 ln t^{2} + 4 : 3 t^{2} - 4 t + 8 arctan (\frac{t}{2}) + 12

= 3 t^{2} - 4 t + 8 arctan (\frac{t}{2}) + 12

Add a constant to the solution

= 3 t^{2} - 4 t + 8 arctan (\frac{t}{2}) + 12 + C

\frac{d}{d x} (e^{4 x} ln (y))

f (x) = 2 x^{3} - 9 x^{2} - 60 x - 2

\int (\frac{x}{x ^{2} + 1}) d x

\int csc^{2} (8 θ) cot (8 θ) d θ

f (θ) = cos (θ^{2})

What is the integral of (6t^3-4t^2+24t)/(t^2+4) ?
The integral of (6t^3-4t^2+24t)/(t^2+4) is 3t^2-4t+8arctan(t/2)+12+C