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

The integral of (12s+12)/((s^2+1)(s-1)^3) is 6arctan(s)+6/(s-1)-6/((s-1)^2)+C

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integral of (12s+12)/((s^2+1)(s-1)^3)

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

6 arctan (s) + \frac{6}{s - 1} - \frac{6}{( s - 1 ) ^{2}} + C

Solution steps

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

Take the partial fraction of

\frac{12 s + 12}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} : \frac{6}{s ^{2} + 1} - \frac{6}{( s - 1 ) ^{2}} + \frac{12}{( s - 1 ) ^{3}}

= \int \frac{6}{s ^{2} + 1} - \frac{6}{( s - 1 ) ^{2}} + \frac{12}{( 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{6}{s ^{2} + 1} d s - \int \frac{6}{( s - 1 ) ^{2}} d s + \int \frac{12}{( s - 1 ) ^{3}} d s

\int \frac{6}{s ^{2} + 1} d s = 6 arctan (s)

\int \frac{6}{( s - 1 ) ^{2}} d s = - \frac{6}{s - 1}

\int \frac{12}{( s - 1 ) ^{3}} d s = - \frac{6}{( s - 1 ) ^{2}}

= 6 arctan (s) - (- \frac{6}{s - 1}) - \frac{6}{( s - 1 ) ^{2}}

Simplify

= 6 arctan (s) + \frac{6}{s - 1} - \frac{6}{( s - 1 ) ^{2}}

Add a constant to the solution

= 6 arctan (s) + \frac{6}{s - 1} - \frac{6}{( s - 1 ) ^{2}} + C

\int \frac{1}{4 - x ^{2}} d x

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

- 6 y^{^{''}} + 24 y = 0

\frac{\partial}{\partial y} (x y e^{x z})

x \to - \infty lim (x^{2} + 2 x)

What is the integral of (12s+12)/((s^2+1)(s-1)^3) ?
The integral of (12s+12)/((s^2+1)(s-1)^3) is 6arctan(s)+6/(s-1)-6/((s-1)^2)+C