# X a k 59 thus p n x x a n 1 dx n k 0 p k n a k x a k

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Chapter 5 / Exercise 5.8
Power System Analysis and Design
Glover/Overbye
Expert Verified
(xa)k.59
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Chapter 5 / Exercise 5.8
Power System Analysis and Design
Glover/Overbye
Expert Verified
ThusPn(x)(xa)n+1dx=nk=0P(k)n(a)k!(xa)kn1dx=nk=0P(k)n(a)k!(xa)kn1dx=n1k=0P(k)n(a)k!·(xa)knkn+P(n)n(a)n!ln|xa|+C.20.We look for the complex roots of the equationx2n=1 = cosπ+isinπ.tag1The roots areαk= cosφk+isinφk,¯αk= cosφkisinφkwherek= 0,1,2, . . . , n1, andφk=(2k+1)π2n. That means, we have exactly 2nrootsβj,j= 1,2, . . . ,2n,which areβ2k+1=αk,β2k+2= ¯αk,fork= 0,1,2, . . . , n. Therefore we haveP(x) :=x2n+ 1 = (xβ1)(xβ2). . .(xβ2n ) =:2nj=1(xβj).Since all the rootsβjare not repeating, there is the following (complex) partial fraction decompositionof1x2n+1:1x2n+ 1=2nj=0Cjxβj=2nj=1Cjm̸=j(xβm)2nm=1(xβm)=2nj=1CjP(x)P(βj)xβjP(x).Consequently,1 =2nj=1CmP(x)P(βj)xβ j.60
In the last equality, by passing to the limitxβm,m= 1,2, . . . ,2n, we obtain1 =Cm·limxβmP(x)P(βm)xβm=Cm·P(βm),i.e.Cm=1P(βm)=122n1m=βm22nm=βm2n.Therefore, we have1x2n+ 1=12n2nj=1βjxβj=12nn1k=0[αkxαk+¯αkx¯αk]=12nn1k=0αk(x¯αk) + ¯αk(xαk)(xαk)(x¯αk)=12nn1k=0x(αk+ ¯αk)2αk¯αkx2(αk+ ¯αk)x+αk¯αk=12nn1k=02xcosφk2x22 cosφk+ 1=12nn1k=0(xcosφk)2 cosφk+ 2 cos2φk2(xcosφk)2+ 1cos2φk=12nn1k=0(xcosφk)2 cosφk2 sin2φk(xcosφk)2+ sin2φk.Thusdxx2n+ 1=12nn1k=0cosφkln(x22xcosφk+ 1)+1nn1k=1sinφkarctan(xcosφksinφk)+C.21.(e): We apply the substitutiont12=x, 12t11dt=dxto get:dx(1 +4x)3x=12t11dt(1 +t3)t4= 12t7dtt3+ 1= 12∫ (t4t131t+ 1+13t+ 1t2t+ 1)dt=125t56t24 ln|t+ 1|+ 4t+ 1t2t+ 1dt=125x5126x164 ln(x112+ 1)+ 2 ln(x16x112+ 1) + 43 arctan(2x11213)+C.61
(g): SinceI:=3(x+ 1)2(x1)4dx=3x+1x1x21dx,we apply the substitutiont3=x+1x1,x=t3+1t31,dx=6t2dt(t31)2andx21 =(t3+ 1)2(t31)21 =4t3(t31)2.So, we getI=t(6t2)dt4t3(t31)2(t31)2=32dt=323x+ 1x1+C.(i): SinceI:=dxn(xa)n+1(xb)n1=nxbxa(xa)(xb)dx,we apply the substitutiontn=xbxa,x=tnabtn1,dx=ntn1(ba)(tn1)2dt, andxa=abtn1,xb=tn(ab)tn1.