What is the integral of (8t^3-3t^2+32t)/(t^2+4) ?

The integral of (8t^3-3t^2+32t)/(t^2+4) is 4t^2-3t+6arctan(t/2)+16+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (8t^3-3t^2+32t)/(t^2+4)

\int \frac{8 t ^{3} - 3 t ^{2} + 32 t}{t ^{2} + 4} d t

4 t^{2} - 3 t + 6 arctan (\frac{t}{2}) + 16 + C

Solution steps

\int \frac{8 t ^{3} - 3 t ^{2} + 32 t}{t ^{2} + 4} d t

Expand

\frac{8 t ^{3} - 3 t ^{2} + 32 t}{t ^{2} + 4} : \frac{8 t ^{3}}{t ^{2} + 4} - \frac{3 t ^{2}}{t ^{2} + 4} + \frac{32 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{8 t ^{3}}{t ^{2} + 4} d t - \int \frac{3 t ^{2}}{t ^{2} + 4} d t + \int \frac{32 t}{t ^{2} + 4} d t

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

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

\int \frac{32 t}{t ^{2} + 4} d t = 16 ln t^{2} + 4

= 4 (t^{2} + 4 - 4 ln t^{2} + 4) - 6 (- arctan (\frac{t}{2}) + \frac{t}{2}) + 16 ln t^{2} + 4

Simplify

4 (t^{2} + 4 - 4 ln t^{2} + 4) - 6 (- arctan (\frac{t}{2}) + \frac{t}{2}) + 16 ln t^{2} + 4 : 4 t^{2} - 3 t + 6 arctan (\frac{t}{2}) + 16

= 4 t^{2} - 3 t + 6 arctan (\frac{t}{2}) + 16

Add a constant to the solution

= 4 t^{2} - 3 t + 6 arctan (\frac{t}{2}) + 16 + C

\frac{\partial}{\partial y} (x^{7 y})

\frac{d}{d x} (ln (- 2 x))

\int - \frac{8 x ^{3}}{( 5 - 2 x ^{4} ) ^{4}} d x

(3 x + 2)^{^{'}}

\frac{d}{d x} (x^{5} + x^{4})

What is the integral of (8t^3-3t^2+32t)/(t^2+4) ?
The integral of (8t^3-3t^2+32t)/(t^2+4) is 4t^2-3t+6arctan(t/2)+16+C