82换元积分法与分部积分法 830k=8-318+30=-8-32 dx d(7-5x) (9) sin 2xsin 3xdx=-lrcos Sx-cos(-x)ldx sinx-isin5x+c (10) ctan (2x dx 1+ sinx (sin 2+ cos 2)2 an(4-2)+c dx d() SinX stnA COs in i ta d(1 (1-x2)2 X (15) 4+x arctan +c 1+(x)2 (6)Jx= In inx+ c (17) 到3=m c
第八章不定积分 (82=(a2 1r(x2-√2)_∫x2±√2) 十c dx (19)x(1-x) (20) tanxdx=/sinr. 1+x= In 11+ I+c dcos In I cosx|+c COS X (21) as xx=(1-sin2xPdsinx=(1-2sin2x+sin'xdsinx E sin X sin 'x+sin 5x+c (2)∫-nk=∫mdk=」 sinxdx t =-In i cosx I+ in I sinx I+c= In i tanx I+c dx de d(x2-3x+8) x2-3x+8 (25)令x+1=t,则x=t-1,dx=dt (t-1)2 (x+1) dt= In I tI 23 In Ix+1+ x+12(x+1) 26. A x=tant,-2<t< 2 dx= asec 'tdt dx dt= sectde
§2换元积分法与分部积分法 In I sect tant l+c= In vx+a2+x I+c =ln1√x2+a2+x|+c (27)令x (x2+a2)2 int +c (28)令x=sint,lt<则dx=aost 「丁mam=m I(1-c0s2t)dcost (1-2c0s2t+ cost)dcost (1-x2)是 -+c (29)令x=tdx=6t5dt 6(5+t2+t2 1+1t+1 11n\t 6x-5x-2x2-6x}-3Lr x+1 (30)令√x+1=t,x=t2-1,dx=2tdt √x+1 t+1 tdt= 2tdt-4 dt x+1+1 203
第八章不定积分 4t+4ln|t+1 1-4√x+1+4ln 2.应用分部积分法求下列不定积分 (1) arcsinxdx; (2) Inxdx (7)[ln(hx)+ _1 (8)(arcsinx)2dx 9)|x23xdx;(10)√x2±adx(a>0); Aw (1)arcsin xdx= xarcsinx-xdarcsinx Arcsin- = xarcsi7x+√1-x2+c (2)Gnxdx xinx-dx= xInx-x+c (3)xcosxdx=x2dsinx =x sinx -2xsinxdx xsinx +2 xdcosx x sinx +2xcosx-2cosxdx x sinx+2xcosx-2sinx +c Lnx inx 1