Jawab:
5.
[tex]\frac{2s^2+s-8}{s^3+4s}[/tex]
[tex]=\frac{2s^2+s-8}{s(s^2+4)}[/tex]
[tex]Mis.\ \frac{2s^2+s-8}{s(s^2+4)} =\frac{A}{s} +\frac{Bs+C}{s^2+4}[/tex]
[tex]\frac{2s^2+s-8}{s(s^2+4)} =\frac{A(s^2+4)+s(Bs+C)}{s(s^2+4)}[/tex]
[tex]\frac{2s^2+s-8}{s(s^2+4)} =\frac{(A+B)s^2+Cs+4A}{s(s^2+4)}[/tex]
setarakan ruas kiri & ruas kanan:
A+B = 2 ... [1]
C = 1 ... [2]
4A = -8 ⇔ A = -2 ... [3]
[1] ⇒ A+B = 2
B = 4
[tex]\int {\frac{2s^2+s-8}{s^3+4s} } \, ds =\int {\frac{-2}{s} } \, ds +\int {\frac{4s+1}{s^2+4} } \, ds[/tex]
[tex]\int {\frac{2s^2+s-8}{s^3+4s} } \, ds =-2(\int {\frac{1}{s} )} \, ds +\int {\frac{4s}{s^2+4} } \, ds+\int {\frac{1}{s^2+4} } \, ds[/tex]
[tex]\int {\frac{2s^2+s-8}{s^3+4s} } \, ds =-2ln(s) +\int {\frac{4s}{s^2+4} } \, \frac{d(s^2+4)}{2s} +\int {\frac{1}{4((\frac{1}{2} s)^2+1)} } \, ds[/tex]
[tex]\int {\frac{2s^2+s-8}{s^3+4s} } \, ds =-2ln(s) +2ln(s^2+4)+\frac{1}{4} \int {\frac{1}{((\frac{1}{2} s)^2+1)} } \, ds[/tex]
[tex]\int {\frac{2s^2+s-8}{s^3+4s} } \, ds =-2ln(s) +2ln(s^2+4)+\frac{1}{4} \times arctan(\frac{1}{2} s)\times2[/tex]
[tex]\int {\frac{2s^2+s-8}{s^3+4s} } \, ds =-2ln(s) +2ln(s^2+4)+\frac{1}{2} arctan(\frac{1}{2} s)+c[/tex]
6.
Mis. u = sin t
[tex]\frac{du}{dt} =cos\ t[/tex]
[tex]dt=\frac{du}{cos\ t}[/tex]
[tex]\int {\frac{cos\ t}{sin^4t-16} } \, dt=\int {\frac{cos\ t}{u^4-16} } \, \frac{du}{cos\ t}[/tex]
[tex]\int {\frac{cos\ t}{sin^4t-16} } \, dt=\int {\frac{1}{u^4-16} } \, du}[/tex]
[tex]Mis.\ \frac{1}{u^4-16} =\frac{Au+B}{u^2+4} +\frac{C}{(u+2)} +\frac{D}{(u-2)}[/tex]
[tex]\frac{1}{u^4-16} =\frac{(Au+B)(u^2-4)+C(u^2+4)(u-2)+D(u^2+4)(u+2)}{(u^2+4)(u+2)(u-2)}[/tex]
[tex]\frac{1}{u^4-16} =\frac{Au^3+Bu^2-4Au-4B+Cu^3-2Cu^2+4Cu-8C+Dx^3+2Du^2+4Du+8D}{(u^2+4)(u+2)(u-2)}[/tex]
[tex]\frac{1}{u^4-16} =\frac{(A+C+D)u^3+(B-2C+2D)u^2+(-4A+4C+4D)u+(-4B-8C+8D)}{(u^2+4)(u+2)(u-2)}[/tex]
dengan cara setarakan seperti di atas:
A+C+D = 0 ... [1]
B-2C+2D =0 ... [2]
-4A+4C+4D = 0 ...[3]
-4B-8C+8D = 1 ... [4]
A = 0
B = -1/8
C = -1/32
D = 1/32
[tex]\int {\frac{1}{u^4-16}} \, du =-\frac{1}{8} \int {\frac{1}{u^2+4}} \, du -\frac{1}{32} \int {\frac{1}{(u+2)} } \, du+\frac{1}{32}\int { \frac{1}{(u-2)}} \, du[/tex]
[tex]\int {\frac{1}{u^4-16}} \, du =-\frac{1}{8}arctan(\frac{u}{2})(\frac{1}{2}) -\frac{1}{32} ln(u+2)} }+\frac{1}{32}ln(u-2)+c[/tex]
[tex]\int {\frac{1}{u^4-16}} \, du =-\frac{1}{16}arctan(\frac{sin\ t}{2}) -\frac{1}{32} ln(sin\ t+2)} }+\frac{1}{32}ln(sin\ t-2)+c[/tex]
[answer.2.content]
