What is the integral of (t^5)/((t^2+4)^2) ?

The integral of (t^5)/((t^2+4)^2) is (t^2)/2-1/4 t^3sin(2arctan(t/2))+t^2-4ln|t^2+4|+4+8/(t^2+4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (t^5)/((t^2+4)^2)

\int \frac{t ^{5}}{( t ^{2} + 4 ) ^{2}} d t

\frac{t ^{2}}{2} - \frac{1}{4} t^{3} sin (2 arctan (\frac{t}{2})) + t^{2} - 4 ln t^{2} + 4 + 4 + \frac{8}{t ^{2} + 4} + C

Solution steps

\int \frac{t ^{5}}{( t ^{2} + 4 ) ^{2}} d t

Long division

\frac{t ^{5}}{( t ^{2} + 4 ) ^{2}} : t + \frac{- 8 t ^{3} - 16 t}{t ^{4} + 8 t ^{2} + 16}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int t d t = \frac{t ^{2}}{2}

\int \frac{- 8 t ^{3} - 16 t}{t ^{4} + 8 t ^{2} + 16} d t = - \frac{1}{4} t^{3} sin (2 arctan (\frac{t}{2})) + t^{2} - 4 ln t^{2} + 4 + 4 + \frac{8}{t ^{2} + 4}

= \frac{t ^{2}}{2} - \frac{1}{4} t^{3} sin (2 arctan (\frac{t}{2})) + t^{2} - 4 ln t^{2} + 4 + 4 + \frac{8}{t ^{2} + 4}

Add a constant to the solution

= \frac{t ^{2}}{2} - \frac{1}{4} t^{3} sin (2 arctan (\frac{t}{2})) + t^{2} - 4 ln t^{2} + 4 + 4 + \frac{8}{t ^{2} + 4} + C

\frac{d}{d x} (ln (arccos (x)))

2 y^{^{''}} + 5 y^{^{'}} - 18 y = 0

\frac{d x}{d t} = t^{\frac{3}{2}}, x (3) = 7

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

\frac{d}{d x} (6 x^{- \frac{1}{2}})

What is the integral of (t^5)/((t^2+4)^2) ?
The integral of (t^5)/((t^2+4)^2) is (t^2)/2-1/4 t^3sin(2arctan(t/2))+t^2-4ln|t^2+4|+4+8/(t^2+4)+C