第三章一元函数积分学(不定积分) 求下列不定积分 解 d In 2.∫cosx+sinx+11+snx (+cos x)- 1+cosx cosx+sinx+11+ sin x 1+sin x, 1+sinx 1(1+sin x 解」(1 1+cosx 1+ cosx 1+cos x 1+cosx 2(1+cosx x3+1) 解,方法一令,_1 dt x3+1)了1(1 In(1+r)+c In 1+ 方法 x(x3+1)x(x3+1) dx l[d(1+x8)-nIx8x)+c=-In(1+8+c 1+ 求下列不定积分 (x+1)2√x2+2x+2 解 (x+1)2√x2+2x+21(x+1)2√(x+1)2+1 tan"t sec t tdt +c sin t sin t x+1 + x 解.令x=tant, coS t dt ds sint r dsin t
第三章 一元函数积分学(不定积分) 一. 求下列不定积分: 1. Ú - + - dx x x x 1 1 ln 1 1 2 解. = - + - Ú dx x x x 1 1 ln 1 1 2 c x x x x d x x ˜ + ¯ ˆ Á Ë Ê - + = - + - + Ú 2 1 1 ln 4 1 1 1 ln 1 1 ln 2 1 2. Ú + + × + + + dx x x x x x 1 cos 1 sin (1 cos ) cos sin 1 2 解. c x x x x d x x dx x x x x x ˜ + ¯ ˆ Á Ë Ê + + = + + + + = + + × + + + Ú Ú 2 2 1 cos 1 sin 2 1 1 cos 1 sin 1 cos 1 sin 1 cos 1 sin (1 cos ) cos sin 1 3. Ú ( +1 ) 8 x x dx 解. 方法一: 令 t x 1 = , t c t t dt dt t t t x x dx = - + + + = - ˜ ¯ ˆ Á Ë Ê + - = + Ú Ú Ú ln(1 ) 8 1 1 1 1 1 1 ( 1 ) 8 8 7 8 2 8 = c x ˜ + ¯ ˆ Á Ë Ê - + 8 1 ln 1 8 1 方法二: Ú Ú Ú + = - - + = + dx x x x x x x dx x x dx ) 1 1 1 ( ( 1 ) ( 1 ) 8 8 7 8 8 7 8 = x x c x d x x dx = - + + + + - Ú Ú ln(1 ) 8 1 ln | | 1 (1 ) 8 1 8 8 8 = c x ˜ + ¯ ˆ Á Ë Ê - + 8 1 ln 1 8 1 二. 求下列不定积分: 1. Ú ( + 1 ) + 2 + 2 2 2 x x x dx 解. Ú Ú + + + + = + + + ( 1) ( 1) 1 ( 1) ( 1) 2 2 2 2 2 2 x x d x x x x dx 令x + 1 = tan t Ú t t t dt tan sec cos 2 2 = Ú + + + + = - + = - c x x x c t t tdt 1 2 2 sin 1 sin cos 2 2 2. Ú + 4 2 x 1 x dx 解. 令 x = tan t, Ú Ú Ú Ú Ú = = = - = - + + + c t t t d t t d t dt t t t t t dt x x dx sin 1 3 sin 1 sin sin sin sin sin cos tan sec cos 1 4 4 2 3 3 4 2 4 2
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:
e+e 解 e +e dx ts∫d (er-e-r)211=arctan(e-e )+c 2(1+4) 解令1=23,a=d t In 2 +4)=](+t)h2in2(-1+r2 tIn 2 In 2 (2+ arctan 2) 四.求下列不定积分 解 99 xd(x-2)-9 x(x-2) 99(x-2 5.4 99xX-2)999×98(x-2)99.9」x(x-2)ax 5.4.3x2 99x-2)998(x-2)99.98.97x-2y99.9897.96(x-2) 5·4·3·2x 5·4·3.2 9098996.9x-2)-9.9.97.96.95(x-2y+c x√1+x tdt 解 令x=1/t 1⊥,4 tan u I rsec udu=- In tanu+ secu+ 五.求下列不定积分: x cos xdx 解.| x cos xdr x(+cos 2x)dx=x d sin 2x x +-xsin 2 in 2xdx
1. Ú - + + dx e e e e x x x x 1 4 2 3 解. Ú Ú Ú = - + - + - = - + + = - + + - - - - - e e c e e d e e dx e e e e dx e e e e x x x x x x x x x x x x x x arctan( ) ( ) 1 ( ) 1 1 4 2 2 2 2 3 2. Ú 2 (1 + 4 ) x x dx 解. 令 x t = 2 , t ln 2 dt dx = c t t dt t t t t dx dt x x ˜ = - - + ¯ ˆ Á Ë Ê + = - + = + Ú Ú Ú ln 2 arctan ln 2 1 1 1 1 ln 2 1 2 (1 4 ) (1 )ln 2 2 2 2 2 = c x x - + + - (2 arctan 2 ) ln 2 1 四. 求下列不定积分: 1. Ú - dx x x 100 5 ( 2 ) 解. Ú Ú Ú - - + - - = - - = - - x x dx x x dx x d x x x 4 99 99 5 5 99 100 5 ( 2 ) 99 5 99 ( 2 ) ( 2 ) 99 1 ( 2 ) = Ú - - × × + ¥ - - - - x x dx x x x x 3 98 98 4 99 5 ( 2 ) 99 98 5 4 99 98 ( 2 ) 5 99 ( 2 ) = 96 2 97 3 98 4 99 5 99 98 97 96 ( 2 ) 5 4 3 99 98 97 ( 2 ) 5 4 99 98 ( 2 ) 5 99( 2 ) × × × - × × - × × - × - × - - - - x x x x x x x x c x x x + × × × × - × × × × - × × × × - × × × - 95 94 99 98 97 96 95 ( 2 ) 5 4 3 2 99 98 97 96 95 ( 2 ) 5 4 3 2 2. Ú + 4 x 1 x dx 解. Ú Ú Ú Ú + = - + = - + - = + 2 2 2 4 4 4 2 4 2 1 ( ) 1 1 1 1 1 1 / 1 t dt t tdt t t t dt t x t x x dx 令 c x x du u u c u u t u + + = - = - + + = - Ú 2 2 4 2 1 ln 2 1 ln | tan sec | 2 1 sec sec 2 1 令 tan 五. 求下列不定积分: 1. Ú x xdx 2 cos 解. Ú Ú Ú x xdx = x + x dx = x + xd sin 2x 4 1 4 1 (1 cos 2 ) 2 1 cos 2 2 Ú = x + x x - sin 2xdx 4 1 sin 2 4 1 4 1 2
x+-xsin 2x +-cOS 2 dx 解∫ sec'xdr=」 j sec xd tan.x= sextans- j tan x sec x tan x =sec x tanx-(sec x-1)sec xdx=sec x tan x+ In sec x+ tan x -sec'xdx 1 xdx=sec x tan x +In sec x tan x +c ∫(mx)d In x) 3(nx)2 (n x)_3(n x)+[ 6Inx dx=_(nx)_3(n-x)_6n x +[6 x x (In x) 3(In x)2 x 6 +c 4. cos(In x ) dx 解∫ cos(In x)dx= x cos(In x)+smmx)= x[cos(In x)+mmx-∫omx)d cos(In x)dx=-[cos(In x)+sin(In x)+c 六.求下列不定积分 x+√1+ tR(x inx+i+xdx= In(x+1+x2yd-I =ln(x+√1+x2) dx √1+ 令x= tant In(r+Ⅵ+x2)1r1 2(1-x2)21-tan2t 2)1 2(1-x2)21-2sin X+ d√2sint 2(1-x2)2√21-2sin2t n(x+√1+x2) 2)4√2 ′oh +c +c 2(1-x2) 1+x2-√2
= x + x x + cos 2x + c 8 1 sin 2 4 1 4 1 2 2. Ú xdx 3 sec 解. Ú Ú Ú sec xdx = sec xd tan x = sec x tan x - tan x sec x tan xdx 3 = Ú Ú x x - x - xdx = x x + x + x - xdx 2 3 sec tan (sec 1 )sec sec tan ln | sec tan | sec xdx = x x + x + x + c Ú ln | sec tan | 2 1 sec tan 2 1 sec3 3. Ú dx x x 2 3 (ln ) 解. Ú Ú Ú = - = - + dx x x x x x dx x d x x 2 2 3 3 2 3 3 (ln ) (ln ) 1 1 (ln ) (ln ) Ú = - - + dx x x x x x x 2 3 2 (ln ) 3 (ln ) 6 ln Ú = - - - + dx x x x x x x x 2 3 2 (ln ) 3 (ln ) 6 ln 6 c x x x x x x x = - - - - + (ln ) 3 (ln ) 6 ln 6 3 2 4. Ú cos(ln x)dx 解. Ú Ú Ú cos(ln x)dx = x cos(ln x ) + sin(ln x )dx = x [cos(ln x ) + sin(ln x )] - cos(ln x )dx \ x x c x x dx = + + Ú [cos(ln ) sin(ln )] 2 cos(ln ) 六. 求下列不定积分: 1. Ú - + + dx x x x x 2 2 2 (1 ) ln( 1 ) 解. Ú Ú - = + + - + + 2 2 2 2 2 1 1 ln( 1 ) 2 1 (1 ) ln( 1 ) x dx x x d x x x x = Ú + × - - - + + dx x x x x x 2 2 2 2 1 1 1 1 2 1 1 1 ln( 1 ) 2 1 令x = tan t tdt x t t x x 2 2 2 2 sec sec 1 1 tan 1 2 1 2(1 ) ln( 1 ) × × - - - + + Ú = dt t t x x x Ú - - - + + 2 2 2 1 2 sin cos 2 1 2(1 ) ln( 1 ) = Ú - - - + + t d t x x x 2 2 2 1 2 sin 2 sin 2 2 1 2(1 ) ln( 1 ) = c t t x x x + - + - - + + 1 2 sin 1 2 sin ln 4 2 1 2(1 ) ln( 1 ) 2 2 = c x x x x x x x + + - + + - - + + 1 2 1 2 ln 4 2 1 2(1 ) ln( 1 ) 2 2 2 2
r arctan x dx arctan 1+x arctan x ∫n-d=Ⅵ+ x?arctan x-m(x+1+x)+c arctan e 解 2/arctane' de= e arctan e"+ -2x e e arctan e"+ - dx dx 1+e 2e(1+e2) an x)+c ln(1+x2)-3 ≥0 七.设f(x)= 求|f(x)dtx (x2+2x-3)ex< ∫(xm1+x2)-3dt 解.」f(x)d ln(1+x2)-[x2-hn(1+x2)]-3 e-2+c1 考虑连续性,所以 x2ln(1+x2)-[x2-ln(1+x2)]-3x+c f(x)x=12 x2+4x+1 0 八.设f(ex)= a sinx+ bcos x,(a,b为不同时为零的常数,求fx) 解.令t=e',x=lnt,f(1) n)+bcos(lnt),所以 f(x)=[asin(In x)+ b cos(In x)]dx la+b)sin(In x)+(b-a)cos(In x)+c 九.求下列不定积分 x=2sin x'v4-x2dx=32 sin t cos'tdt =-32 (1-cos ()cos ld cost
2. Ú + dx x x x 2 1 arctan 解. Ú Ú Ú + + = + = + - + dx x x dx xd x x x x x x 2 2 2 2 2 1 1 arctan 1 1 arctan 1 arctan = dx x x x x c x x x = + - + + + + + - Ú 1 arctan ln( 1 ) 1 1 1 arctan 2 2 2 2 3. Ú dx e e x x 2 arctan 解. dx e e dx e de e e e e e x x x x x x x x x Ú Ú Ú + = - = - + - - - 2 2 2 2 2 2 1 1 arctan 2 1 arctan 2 arctan 1 dx e e e e x x x x Ú + = - + - - 2 2 2 1 1 arctan 2 1 Ú + = - + - dx e e e e x x x x (1 ) 1 2 1 arctan 2 1 2 2 dx e e e x c e e e e e x x x x x x x x = - + + + + = - + - - - - Ú ( arctan arctan ) 2 1 ) 1 1 ( 2 1 arctan 2 1 2 2 2 七. 设 Ó Ì Ï + - + - = -x x x e x x f x ( 2 3 ) ln(1 ) 3 ( ) 2 2 0 0 < ³ x x , 求 Ú f (x )dx . 解. Ô Ó Ô Ì Ï + - + - = - Ú Ú Ú x x e dx x x dx f x dx x ( 2 3) ( ln(1 ) 3) ( ) 2 2 Ô Ó Ô Ì Ï - + + + + - - + - + = - 1 2 2 2 2 2 ( 4 1 ) [ ln(1 )] 3 2 1 ln(1 ) 2 1 x x e c x x x x x c x 0 0 < ³ x x 考虑连续性, 所以 c =－1+ c1, c1 = 1 + c Ú f (x )dx Ô Ó Ô Ì Ï - + + + + + - - + - + = - x x e c x x x x x c x ( 4 1 ) 1 [ ln(1 )] 3 2 1 ln(1 ) 2 1 2 2 2 2 2 0 0 < ³ x x 八. 设 f e a x b x x '( ) = sin + cos , (a, b 为不同时为零的常数), 求 f(x). 解. 令t e x t x = ， = ln , f '(t ) = a sin(ln t ) + b cos(ln t ) , 所以 Ú f (x ) = [a sin(ln x ) + b cos(ln x )]dx = a b x b a x c x [( + )sin(ln ) + ( - ) cos(ln )] + 2 九. 求下列不定积分: 1. Ú x - x dx 3 2 4 解. 令 x = 2sin t Ú Ú Ú x 4 - x dx = 32 sin t cos tdt = - 32 (1 - cos t) cos td cos t 3 2 3 2 2 2