What is the integral of (20s+20)/((s^2+1)(s-1)^3) ?

The integral of (20s+20)/((s^2+1)(s-1)^3) is 10arctan(s)+(10)/(s-1)-(10)/((s-1)^2)+C

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integral of (20s+20)/((s^2+1)(s-1)^3)

\int \frac{20 s + 20}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

10 arctan (s) + \frac{10}{s - 1} - \frac{10}{( s - 1 ) ^{2}} + C

Solution steps

\int \frac{20 s + 20}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

Take the partial fraction of

\frac{20 s + 20}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} : \frac{10}{s ^{2} + 1} - \frac{10}{( s - 1 ) ^{2}} + \frac{20}{( s - 1 ) ^{3}}

= \int \frac{10}{s ^{2} + 1} - \frac{10}{( s - 1 ) ^{2}} + \frac{20}{( 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{10}{s ^{2} + 1} d s - \int \frac{10}{( s - 1 ) ^{2}} d s + \int \frac{20}{( s - 1 ) ^{3}} d s

\int \frac{10}{s ^{2} + 1} d s = 10 arctan (s)

\int \frac{10}{( s - 1 ) ^{2}} d s = - \frac{10}{s - 1}

\int \frac{20}{( s - 1 ) ^{3}} d s = - \frac{10}{( s - 1 ) ^{2}}

= 10 arctan (s) - (- \frac{10}{s - 1}) - \frac{10}{( s - 1 ) ^{2}}

Simplify

= 10 arctan (s) + \frac{10}{s - 1} - \frac{10}{( s - 1 ) ^{2}}

Add a constant to the solution

= 10 arctan (s) + \frac{10}{s - 1} - \frac{10}{( s - 1 ) ^{2}} + C

\frac{d}{d x} ((3 - 4 x)^{\frac{4}{3}})

f (x) = [(x - 2)^{2} - x]^{2}, at x = 3

4 - x^{2}, y = 0

\int_{3}^{6} 10000 e^{0.02 t} d t

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

What is the integral of (20s+20)/((s^2+1)(s-1)^3) ?
The integral of (20s+20)/((s^2+1)(s-1)^3) is 10arctan(s)+(10)/(s-1)-(10)/((s-1)^2)+C