Respuesta :
Solution of integral [tex]\int \frac{ds}{s^{2}(s-1)^{2} }[/tex] is [tex]2ln|\frac{s}{s-1}|-\frac{1}{s}-\frac{1}{s-1}+c[/tex].
We have to evaluate [tex]\int \frac{ds}{s^{2}(s-1)^{2} }[/tex]
By using partial fractions, evaluate the integral
Let [tex]\frac{1}{s^{2}(s-1)^{2} } =\frac{A}{s} +\frac{B}{s^{2}} +\frac{c}{s-1} +\frac{D}{(s-1)^{2} }[/tex]
[tex]1=As(s-1)^{2} +B(s-1)^{2} +Cs^{2}(s-1)+Ds^{2}[/tex]
Put s = 0, 1 = B.1 ⇒ B = 1
Put s= 1, 1 = D ⇒ D = 1
Put s = -1, [tex]1=A(-1)(-2)^{2} +B(-2)^{2} +C(-1)^{2} (-2)+D(-1)^{2}[/tex]
1 = -4A + 4B -2C + D
1 = -4A + 4 -2C + 1
-4 = -4A -2C
2 = 2A + C----------------(1)
Put s = 2, 1 = A(2) + B +C(4) +D(4)
1 = 2A + 1 + 4C +4
-4 = 2A + 4C
-2 = A + 2C---------------(2)
From(1),
C = 2 -2A ---------------(3)
From (2),
⇒ -2 = A + 4 -4A
⇒ -6 = -3A ⇒ A = 2
From (3)
C = 2 - 2(2)
⇒ C = -2
Thus [tex]\int \frac{ds}{s^{2}(s-1)^{2} }= \int (\frac{2}{s} +\frac{1}{s^{2} }- \frac{2}{s-1} +\frac{1}{(s-1)^{2} } )ds[/tex]
= [tex]2 ln|s|-\frac{1}{s} -2ln|s-1|-\frac{1}{s-1}+c[/tex]
= [tex]2ln|\frac{s}{s-1}|-\frac{1}{s}-\frac{1}{s-1}+c[/tex]
Thus,
Solution of integral [tex]\int \frac{ds}{s^{2}(s-1)^{2} }[/tex] is [tex]2ln|\frac{s}{s-1}|-\frac{1}{s}-\frac{1}{s-1}+c[/tex].
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