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

The integral of ()/(s^2(s-1)^2) is (ds)/((-1+s)^2s^2)x+C

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integral of ()/(s^2(s-1)^2)

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

\frac{d s}{( - 1 + s ) ^{2} s ^{2}} x + C

Solution steps

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

Integral of a constant:

\int a d x = a x

= \frac{d s}{( - 1 + s ) ^{2} s ^{2}} x

Add a constant to the solution

= \frac{d s}{( - 1 + s ) ^{2} s ^{2}} x + C

\frac{d}{d x} (\frac{2}{x} - \frac{3 x}{3})

\frac{d}{d x} (\frac{x ^{3}}{4} + \frac{1}{3 x})

\int 2 d x

x \to 1 lim ([5 f (x) + g (x)])

\frac{d}{d x} (ln (e x))

What is the integral of ()/(s^2(s-1)^2) ?
The integral of ()/(s^2(s-1)^2) is (ds)/((-1+s)^2s^2)x+C