What is the integral of (4t^3-t^2+16t)/(t^2+4) ?

The integral of (4t^3-t^2+16t)/(t^2+4) is 2t^2-t+2arctan(t/2)+8+C

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integral of (4t^3-t^2+16t)/(t^2+4)

\int \frac{4 t ^{3} - t ^{2} + 16 t}{t ^{2} + 4} d t

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

Solution steps

\int \frac{4 t ^{3} - t ^{2} + 16 t}{t ^{2} + 4} d t

Expand

\frac{4 t ^{3} - t ^{2} + 16 t}{t ^{2} + 4} : \frac{4 t ^{3}}{t ^{2} + 4} - \frac{t ^{2}}{t ^{2} + 4} + \frac{16 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{4 t ^{3}}{t ^{2} + 4} d t - \int \frac{t ^{2}}{t ^{2} + 4} d t + \int \frac{16 t}{t ^{2} + 4} d t

\int \frac{4 t ^{3}}{t ^{2} + 4} d t = 2 (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{16 t}{t ^{2} + 4} d t = 8 ln t^{2} + 4

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

Simplify

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{n}{n ^{2} + 3}

\int (x y^{2} - cos (x) sin (x)) d x

y^{^{'}} - 2 x y = 8 x

\int (x^{4} + sec^{2} (x)) d x

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

What is the integral of (4t^3-t^2+16t)/(t^2+4) ?
The integral of (4t^3-t^2+16t)/(t^2+4) is 2t^2-t+2arctan(t/2)+8+C