l(√1+x dx +x 解.令x=tant ec t coS t d sin t dt dt 2x2+1)√1+ 2 tant+1)sect J2sin2t+cost arctan sin t+c= arctan +c 解.令x= a sin t dx ,2 sin 2t a cos tdt 2( 1-cos2t dt2q't--a sin 2t+c a cost arcsin 解.令x=sint ∫√-x)dk=」osth=∫ (1+ CoS20).[1+2 cos 2t+cos" 2t I+sin 2t+l(1+cos 4n)dt=t+sin 2t+asin 4t+c --arcsin x+sin 2t(1+cos 20)+c 4+1 arcsin x +-2 sin t cost( )+C =Arcsin x+-xV1-x2(5-2x2)+c x dx =-m全=sm-mosm cos u+c 求下列不定积分= c x x x x + + + ˜ ˜ ¯ ˆ Á Á Ë Ê + - 2 3 2 1 1 3 1 3. Ú + + 2 2 (2 x 1 ) 1 x dx 解. 令 x = tan t Ú Ú Ú Ú + = + = + = + + t d t dt t t t dt t t t x x dx 2 2 2 2 2 2 2 1 sin sin 2 sin cos cos (2 tan 1)sec sec (2 1) 1 = c x x t c + + + = 2 1 arctan sin arctan 4. Ú - 2 2 2 a x x dx (a > 0) 解. 令 x = a sin t Ú Ú Ú = - + - = × = - dt a t a t c t a a t a t a tdt a x x dx sin 2 4 1 2 1 2 1 cos2 cos sin cos 2 2 2 2 2 2 2 2 = a x c a x a a x ˜ + ¯ ˆ Á Ë Ê - - 2 2 2 2 arcsin 2 5. Ú - x dx 2 3 (1 ) 解. 令 x = sin t Ú Ú Ú Ú + + = + - = = dt t t dt t x dx tdt 4 1 2 cos2 cos 2 4 (1 cos2 ) (1 ) cos 2 2 2 3 4 = Ú t + t + + t dt = t + t + sin 4t + c 32 1 sin 2 4 1 8 3 (1 cos 4 ) 8 1 sin 2 4 1 4 1 = x + t + cos 2t) + c 4 1 sin 2 (1 4 1 arcsin 8 3 = c t x t t + + - + ) 4 4 1 2 sin 2 sin cos ( 4 1 arcsin 8 3 2 = x + x 1- x (5 - 2x ) + c 8 1 arcsin 8 3 2 2 6. Ú - dx x x 4 2 1 解. 令 t x 1 = Ú Ú Ú ˜ = - - ¯ ˆ Á Ë Ê - - = - dt t t dt t t t t dx x x 2 2 4 2 2 4 2 1 1 1 1 1 令t = sin u Ú - u udu 2 sin cos = c x x u c + - + = 3 2 3 3 3 ( 1 ) cos 3 1 三. 求下列不定积分: