求下列不定积分(其中α＞0)：
(1)∫dx/(√2x-1)+1
(2)∫dx/[x/(x8)]；
(3)∫dx/(α2-x2)3/2；
(4)∫dx/(α2+x2)3/2；
(5)∫dx/(x2-α2)3/2；
(6)∫(x2/√α2-x2)dx
(7)∫(√x2-α2/x2)dx
(8)(√x2+α2/x4)dx；
(9)∫dx/(x2√x2-α2)；
(10)∫dx/(x2√α2-x2
(1)令√2x-1=t，则x=(t2+1)/2(x≥1/2)，dx=tdt ∫dx/(√2x-1)+1 =∫tdt/(t+1)=∫[1-1/(t+1)]dt =t-ln(t+1)+C =√2x-1-ln(√2x-1+1)+C (2)令x=1/t，则dx=-(1/t2)dt ∫dx/[x(1+x8) =∫{[-(1/t2)]/[1/t(1+1/t8)]}dt =-∫[t7/(t8+1)]dt =-(1/8)∫[d(t8+1)]/(t8+1) =-(1/8)ln(t8+1)+C =-(1/8)ln[(1+x8)/x8]+C =1n|x|-(1/8)ln(1+x8)+C (3)令x=αsint(0＜t＜π/2)，则dx=αcotdt ∫dx/(α2-x2)3/2 =∫[αcost/(α3•cos3t)]dt =1/α2∫sec2tdt =1/α2tant+C =1/α2(x/√α2-x2)+C (4)令x=αtant(-(π/2)＜x＜π/2)，则dx=αsec2tdt 则∫dx/(α2+x2)3/2 =∫αsec2tdt/α3sec3t =1/α2∫(1/sect)dt =1/α2∫costdt =(1/α2)sint+C =[x/(α2√α2+x)+C (5)令x=αsect(0＜t＜π/2)，则dx=αsecttantdt ∫dx/(x2-α2)3/2 =∫(αsecttant/α3tan3t) =1/α∫(cost/sin2t) =1/α2∫(dsint/sin2t) =-(1/α2)•1/sint+C =-(1/α2)•(x/√x2-α2)+C =-[x/(α2√x2-α2)=+C (6)令x=αsint(0＜t＜π/2)，则dx=αcotdt ∫(x2/√α2-x2)dx =∫(α2sin2t/αcost)•αcostdt =α2∫p(1-cos2t)/2]dt =α2t/2-α2sin2t/4+C =(α2/2)arcsin(x/α)-x/2√α2-x2+C (7)令x=αsect(0＜t＜π/2)，则dx=αsecttanntdt2-α2/x2)dx =∫αtant/α2sec2t•αsect•tantdt =∫sect-costdt． =In|sect+tant|-sint+C =ln|1/α+√x2-α2/α|-√x2-α2/x+c (8)令x=αtant(0＜t＜π/2)，则dx=αsec2tdt ∫(√x2-α2/x4)dx =∫αsect/α4tan4t•αsevtdt =1/α2∫(1/sint)dsint =-1/3α2sin3t+C =-(1/3α2)[(√α2+x2)3/x3]+C (9)令x=αsect(0＜t＜π/2)，则dx=αsecttantdt ∫dx/(x2√x2-α2) =∫[(αsect•tant)/(α2sec2t•αtant)]dt =1/α2∫cotdt =(1/α2)sint+C =√x2-α2/α2x+C (10)令x=αsint(0＜t＜π/2) ∫=dx/(x2√αx2-xx2) =∫[αcost/(αx2sinx2t•cost)]dt =1/αx2∫csctdt． =-(cott/α2+C =-(√αx2-xx2/αx2x)+C
