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

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

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

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

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

Solution steps

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

Expand

\frac{6 t ^{3} - t ^{2} + 24 t}{t ^{2} + 4} : \frac{6 t ^{3}}{t ^{2} + 4} - \frac{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{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{t ^{2}}{t ^{2} + 4} d t = - 2 arctan (\frac{t}{2}) + t

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

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

Simplify

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

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

Add a constant to the solution

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

- \frac{5 s}{( ( s + 1 ) ^{2} + 3 )}

\frac{d}{d x} (e^{x^{3}} y)

\int 10 x^{\frac{3}{2}} + 6 x - 8 d x

\int \frac{5}{x} + sec^{2} (x) d x

x \to 1 lim (\frac{- 3 x}{x - 1})

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