What is the integral of (16s+16)/((s^2+1)(s-1)^3) ?

The integral of (16s+16)/((s^2+1)(s-1)^3) is 8arctan(s)+8/(s-1)-8/((s-1)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (16s+16)/((s^2+1)(s-1)^3)

\int \frac{16 s + 16}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

8 arctan (s) + \frac{8}{s - 1} - \frac{8}{( s - 1 ) ^{2}} + C

Solution steps

\int \frac{16 s + 16}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

Take the partial fraction of

\frac{16 s + 16}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} : \frac{8}{s ^{2} + 1} - \frac{8}{( s - 1 ) ^{2}} + \frac{16}{( s - 1 ) ^{3}}

= \int \frac{8}{s ^{2} + 1} - \frac{8}{( s - 1 ) ^{2}} + \frac{16}{( s - 1 ) ^{3}} d s

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}{s ^{2} + 1} d s - \int \frac{8}{( s - 1 ) ^{2}} d s + \int \frac{16}{( s - 1 ) ^{3}} d s

\int \frac{8}{s ^{2} + 1} d s = 8 arctan (s)

\int \frac{8}{( s - 1 ) ^{2}} d s = - \frac{8}{s - 1}

\int \frac{16}{( s - 1 ) ^{3}} d s = - \frac{8}{( s - 1 ) ^{2}}

= 8 arctan (s) - (- \frac{8}{s - 1}) - \frac{8}{( s - 1 ) ^{2}}

Simplify

= 8 arctan (s) + \frac{8}{s - 1} - \frac{8}{( s - 1 ) ^{2}}

Add a constant to the solution

= 8 arctan (s) + \frac{8}{s - 1} - \frac{8}{( s - 1 ) ^{2}} + C

x \to 0.999 lim (x^{2} + 2)

\frac{2 s}{s ^{2} + 6}

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

\frac{d y}{d t} = \frac{8}{7} y

\frac{\partial}{\partial x} ((2 x + 3) (y - 2))

What is the integral of (16s+16)/((s^2+1)(s-1)^3) ?
The integral of (16s+16)/((s^2+1)(s-1)^3) is 8arctan(s)+8/(s-1)-8/((s-1)^2)+C