What is the integral of (28s+28)/((s^2+1)(s-1)^3) ?

The integral of (28s+28)/((s^2+1)(s-1)^3) is 14arctan(s)+(14)/(s-1)-(14)/((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 (28s+28)/((s^2+1)(s-1)^3)

\int \frac{28 s + 28}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

14 arctan (s) + \frac{14}{s - 1} - \frac{14}{( s - 1 ) ^{2}} + C

Solution steps

\int \frac{28 s + 28}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

Take the partial fraction of

\frac{28 s + 28}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} : \frac{14}{s ^{2} + 1} - \frac{14}{( s - 1 ) ^{2}} + \frac{28}{( s - 1 ) ^{3}}

= \int \frac{14}{s ^{2} + 1} - \frac{14}{( s - 1 ) ^{2}} + \frac{28}{( 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{14}{s ^{2} + 1} d s - \int \frac{14}{( s - 1 ) ^{2}} d s + \int \frac{28}{( s - 1 ) ^{3}} d s

\int \frac{14}{s ^{2} + 1} d s = 14 arctan (s)

\int \frac{14}{( s - 1 ) ^{2}} d s = - \frac{14}{s - 1}

\int \frac{28}{( s - 1 ) ^{3}} d s = - \frac{14}{( s - 1 ) ^{2}}

= 14 arctan (s) - (- \frac{14}{s - 1}) - \frac{14}{( s - 1 ) ^{2}}

Simplify

= 14 arctan (s) + \frac{14}{s - 1} - \frac{14}{( s - 1 ) ^{2}}

Add a constant to the solution

= 14 arctan (s) + \frac{14}{s - 1} - \frac{14}{( s - 1 ) ^{2}} + C

\frac{\partial}{\partial x} (3 x + 4 y^{2} + 3 x^{2} y^{3})

\int_{1}^{3} u^{2} + b^{2} d u

\int x - 2 π d x

\int - t^{2} e^{- t} d t

\frac{\partial}{\partial x} (arctan (x^{3} y^{2}))

What is the integral of (28s+28)/((s^2+1)(s-1)^3) ?
The integral of (28s+28)/((s^2+1)(s-1)^3) is 14arctan(s)+(14)/(s-1)-(14)/((s-1)^2)+C