What is the integral of (24s+24)/((s^2+1)(s-1)^3) ?

The integral of (24s+24)/((s^2+1)(s-1)^3) is 12arctan(s)+(12)/(s-1)-(12)/((s-1)^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (24s+24)/((s^2+1)(s-1)^3)

\int \frac{24 s + 24}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

12 arctan (s) + \frac{12}{s - 1} - \frac{12}{( s - 1 ) ^{2}} + C

Solution steps

\int \frac{24 s + 24}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

Take the partial fraction of

\frac{24 s + 24}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} : \frac{12}{s ^{2} + 1} - \frac{12}{( s - 1 ) ^{2}} + \frac{24}{( s - 1 ) ^{3}}

= \int \frac{12}{s ^{2} + 1} - \frac{12}{( s - 1 ) ^{2}} + \frac{24}{( 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{12}{s ^{2} + 1} d s - \int \frac{12}{( s - 1 ) ^{2}} d s + \int \frac{24}{( s - 1 ) ^{3}} d s

\int \frac{12}{s ^{2} + 1} d s = 12 arctan (s)

\int \frac{12}{( s - 1 ) ^{2}} d s = - \frac{12}{s - 1}

\int \frac{24}{( s - 1 ) ^{3}} d s = - \frac{12}{( s - 1 ) ^{2}}

= 12 arctan (s) - (- \frac{12}{s - 1}) - \frac{12}{( s - 1 ) ^{2}}

Simplify

= 12 arctan (s) + \frac{12}{s - 1} - \frac{12}{( s - 1 ) ^{2}}

Add a constant to the solution

= 12 arctan (s) + \frac{12}{s - 1} - \frac{12}{( s - 1 ) ^{2}} + C

What is the integral of (24s+24)/((s^2+1)(s-1)^3) ?
The integral of (24s+24)/((s^2+1)(s-1)^3) is 12arctan(s)+(12)/(s-1)-(12)/((s-1)^2)+C