What is the integral of (18s+18)/((s^2+1)(s-1)^3) ?

The integral of (18s+18)/((s^2+1)(s-1)^3) is 9arctan(s)+9/(s-1)-9/((s-1)^2)+C

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integral of (18s+18)/((s^2+1)(s-1)^3)

\int \frac{18 s + 18}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

9 arctan (s) + \frac{9}{s - 1} - \frac{9}{( s - 1 ) ^{2}} + C

Solution steps

\int \frac{18 s + 18}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

Take the partial fraction of

\frac{18 s + 18}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} : \frac{9}{s ^{2} + 1} - \frac{9}{( s - 1 ) ^{2}} + \frac{18}{( s - 1 ) ^{3}}

= \int \frac{9}{s ^{2} + 1} - \frac{9}{( s - 1 ) ^{2}} + \frac{18}{( 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{9}{s ^{2} + 1} d s - \int \frac{9}{( s - 1 ) ^{2}} d s + \int \frac{18}{( s - 1 ) ^{3}} d s

\int \frac{9}{s ^{2} + 1} d s = 9 arctan (s)

\int \frac{9}{( s - 1 ) ^{2}} d s = - \frac{9}{s - 1}

\int \frac{18}{( s - 1 ) ^{3}} d s = - \frac{9}{( s - 1 ) ^{2}}

= 9 arctan (s) - (- \frac{9}{s - 1}) - \frac{9}{( s - 1 ) ^{2}}

Simplify

= 9 arctan (s) + \frac{9}{s - 1} - \frac{9}{( s - 1 ) ^{2}}

Add a constant to the solution

= 9 arctan (s) + \frac{9}{s - 1} - \frac{9}{( s - 1 ) ^{2}} + C

What is the integral of (18s+18)/((s^2+1)(s-1)^3) ?
The integral of (18s+18)/((s^2+1)(s-1)^3) is 9arctan(s)+9/(s-1)-9/((s-1)^2)+C