高等数学(第七版)同济大学 总习题四(前半部分) 个人解答
高等数学(第七版)同济大学 总习题四(前半部分)
1. 填空: \begin{aligned}&1. \ 填空:&\end{aligned} 1. 填空:
( 1 ) ∫ x 3 e x d x = _ _ _ _ _ _ _ _ ( 2 ) ∫ x + 5 x 2 − 6 x + 13 d x = _ _ _ _ _ _ _ _ \begin{aligned} &\ \ (1)\ \ \int x^3e^xdx=\_\_\_\_\_\_\_\_ \\\\ &\ \ (2)\ \ \int \frac{x+5}{x^2-6x+13}dx=\_\_\_\_\_\_\_\_ & \end{aligned} (1) ∫x3exdx=________ (2) ∫x2−6x+13x+5dx=________
解:
( 1 ) 设 u = x 3 , d v = e x d x ,则 d u = 3 x 2 d x , v = e x ,得 ∫ x 3 e x d x = x 3 e x − 3 ∫ x 2 e x d x ,设 u = x 2 , d v = e x d x ,则 d u = 2 x d x , v = e x ,得 ∫ x 3 e x d x = x 3 e x − 3 ∫ x 2 e x d x = x 3 e x − 3 ( x 2 e x − 2 ∫ x e x d x ) ,设 u = x , d v = e x d x , 则 d u = d x , v = e x ,得 ∫ x 3 e x d x = x 3 e x − 3 ∫ x 2 e x d x = x 3 e x − 3 ( x 2 e x − 2 ∫ x e x d x ) = x 3 e x − 3 ( x 2 e x − 2 ∫ x e x d x ) = x 3 e x − 3 x 2 e x + 6 x e x − 6 e x + C = e x ( x 3 − 3 x 2 + 6 x − 6 ) + C ( 2 ) ∫ x + 5 x 2 − 6 x + 13 d x = ∫ x − 3 ( x − 3 ) 2 + 4 d ( x − 3 ) + 8 ∫ 1 ( x − 3 ) 2 + 4 d ( x − 3 ) ,根据积分表公式 23 和 19 ,得 ∫ x − 3 ( x − 3 ) 2 + 4 d ( x − 3 ) + 8 ∫ 1 ( x − 3 ) 2 + 4 d ( x − 3 ) = 1 2 l n ( x 2 − 6 x + 13 ) + 4 a r c t a n x − 3 2 + C \begin{aligned} &\ \ (1)\ 设u=x^3,dv=e^xdx,则du=3x^2dx,v=e^x,得\\\\ &\ \ \ \ \ \ \ \ \ \int x^3e^xdx=x^3e^x-3\int x^2e^xdx,设u=x^2,dv=e^xdx,则du=2xdx,v=e^x,得\\\\ &\ \ \ \ \ \ \ \ \ \int x^3e^xdx=x^3e^x-3\int x^2e^xdx=x^3e^x-3(x^2e^x-2\int xe^xdx),设u=x,dv=e^xdx,\\\\ &\ \ \ \ \ \ \ \ \ \ 则du=dx,v=e^x,得\int x^3e^xdx=x^3e^x-3\int x^2e^xdx=x^3e^x-3(x^2e^x-2\int xe^xdx)=\\\\ &\ \ \ \ \ \ \ \ \ \ x^3e^x-3(x^2e^x-2\int xe^xdx)= x^3e^x-3x^2e^x+6xe^x-6e^x+C=e^x(x^3-3x^2+6x-6)+C\\\\ &\ \ (2)\ \int \frac{x+5}{x^2-6x+13}dx=\int \frac{x-3}{(x-3)^2+4}d(x-3)+8\int \frac{1}{(x-3)^2+4}d(x-3),根据积分表公式23和19,得\\\\ &\ \ \ \ \ \ \ \ \int \frac{x-3}{(x-3)^2+4}d(x-3)+8\int \frac{1}{(x-3)^2+4}d(x-3)=\frac{1}{2}ln(x^2-6x+13)+4arctan\ \frac{x-3}{2}+C & \end{aligned} (1) 设u=x3,dv=exdx,则du=3x2dx,v=ex,得 ∫x3exdx=x3ex−3∫x2exdx,设u=x2,dv=exdx,则du=2xdx,v=ex,得 ∫x3exdx=x3ex−3∫x2exdx=x3ex−3(x2ex−2∫xexdx),设u=x,dv=exdx, 则du=dx,v=ex,得∫x3exdx=x3ex−3∫x2exdx=x3ex−3(x2ex−2∫xexdx)= x3ex−3(x2ex−2∫xexdx)=x3ex−3x2ex+6xex−6ex+C=ex(x3−3x2+6x−6)+C (2) ∫x2−6x+13x+5dx=∫(x−3)2+4x−3d(x−3)+8∫(x−3)2+41d(x−3),根据积分表公式23和19,得 ∫(x−3)2+4x−3d(x−3)+8∫(x−3)2+41d(x−3)=21ln(x2−6x+13)+4arctan 2x−3+C
2. 以下两题中给出了四个结论,从中选出一个正确的结论: \begin{aligned}&2. \ 以下两题中给出了四个结论,从中选出一个正确的结论:&\end{aligned} 2. 以下两题中给出了四个结论,从中选出一个正确的结论:
( 1 ) 已知 f ′ ( x ) = 1 x ( 1 + 2 l n x ) ,且 f ( 1 ) = 1 ,则 f ( x ) 等于 ( ) : ( A ) l n ( 1 + 2 l n x ) + 1 ( B ) 1 2 l n ( 1 + 2 l n x ) + 1 ( C ) 1 2 l n ( 1 + 2 l n x ) + 1 2 ( D ) 2 l n ( 1 + 2 l n x ) + 1 ( 2 ) 在下列等式中,正确的结果是 ( ) . ( A ) ∫ f ′ ( x ) d x = f ( x ) ( B ) ∫ d f ( x ) = f ( x ) ( C ) d d x ∫ f ( x ) d x = f ( x ) ( D ) d ∫ f ( x ) = f ( x ) \begin{aligned} &\ \ (1)\ 已知f'(x)=\frac{1}{x(1+2ln\ x)},且f(1)=1,则f(x)等于(\ \ \ \ ):\\\\ &\ \ (A)\ ln(1+2ln\ x)+1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (B)\ \frac{1}{2}ln(1+2ln\ x)+1\\\\ &\ \ (C)\ \frac{1}{2}ln(1+2ln\ x)+\frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (D)\ 2ln(1+2ln\ x)+1\\\\ &\ \ (2)\ 在下列等式中,正确的结果是(\ \ \ \ ).\\\\ &\ \ (A)\ \int f'(x)dx=f(x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (B)\ \int df(x)=f(x)\\\\ &\ \ (C)\ \frac{d}{dx}\int f(x)dx=f(x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (D)\ d\int f(x)=f(x) & \end{aligned} (1) 已知f′(x)=x(1+2ln x)1,且f(1)=1,则f(x)等于( ): (A) ln(1+2ln x)+1 (B) 21ln(1+2ln x)+1 (C) 21ln(1+2ln x)+21 (D) 2ln(1+2ln x)+1 (2) 在下列等式中,正确的结果是( ). (A) ∫f′(x)dx=f(x) (B) ∫df(x)=f(x) (C) dxd∫f(x)dx=f(x) (D) d∫f(x)=f(x)
解:
( 1 ) 求 ∫ 1 x ( 1 + 2 l n x ) d x ,设 u = l n x ,则 x = e u , d x = e u d u ,得 ∫ 1 x ( 1 + 2 l n x ) d x = ∫ e u e u + 2 u e u d u = 1 2 ∫ 1 1 + 2 u d ( 1 + 2 u ) = 1 2 l n ∣ 1 + 2 l n x ∣ + C , 因为 f ( 1 ) = 1 ,确定 C = 1 ,选 B ( 2 ) ∫ d f ( x ) = ∫ f ′ ( x ) d x = f ( x ) + C , d d x ∫ f ( x ) d x = f ( x ) , d ∫ f ( x ) d x = ( d d x ∫ f ( x ) d x ) d x = f ( x ) d x ,所以选 C \begin{aligned} &\ \ (1)\ 求\int \frac{1}{x(1+2ln\ x)}dx,设u=ln\ x,则x=e^u,dx=e^udu,得\\\\ &\ \ \ \ \ \ \ \ \ \ \ \ \int \frac{1}{x(1+2ln\ x)}dx=\int \frac{e^u}{e^u+2ue^u}du=\frac{1}{2}\int \frac{1}{1+2u}d(1+2u)=\frac{1}{2}ln\ |1+2ln\ x|+C,\\\\ &\ \ \ \ \ \ \ \ \ \ \ \ 因为f(1)=1,确定C=1,选B\\\\ &\ \ (2)\ \int df(x)=\int f'(x)dx=f(x)+C,\frac{d}{dx}\int f(x)dx=f(x),\\\\ &\ \ \ \ \ \ \ \ \ d\int f(x)dx=\left(\frac{d}{dx}\int f(x)dx\right)dx=f(x)dx,所以选C & \end{aligned} (1) 求∫x(1+2ln x)1dx,设u=ln x,则x=eu,dx=eudu,得 ∫x(1+2ln x)1dx=∫eu+2ueueudu=21∫1+2u1d(1+2u)=21ln ∣1+2ln x∣+C, 因为f(1)=1,确定C=1,选B (2) ∫df(x)=∫f′(x)dx=f(x)+C,dxd∫f(x)dx=f(x), d∫f(x)dx=(dxd∫f(x)dx)dx=f(x)dx,所以选C
3. 已知 s i n x x 是 f ( x ) 的一个原函数,求 ∫ x 3 f ′ ( x ) d x . \begin{aligned}&3. \ 已知\frac{sin\ x}{x}是f(x)的一个原函数,求\int x^3f'(x)dx.&\end{aligned} 3. 已知xsin x是f(x)的一个原函数,求∫x3f′(x)dx.
解:
设 F ( x ) = s i n x x ,则 f ( x ) = F ′ ( x ) = ( s i n x x ) ′ = x c o s x − s i n x x 2 = c o s x x − s i n x x 2 , 则 f ′ ( x ) = − x s i n x − c o s x x 2 − x 2 c o s x − 2 x s i n x x 4 = − ( x 2 − 2 ) s i n x + 2 x c o s x x 3 , ∫ x 3 f ′ ( x ) d x = − ∫ ( ( x 2 − 2 ) s i n x + 2 x c o s x ) d x = − ∫ x 2 s i n x d x + 2 ∫ s i n x d x − 2 ∫ x c o s x d x = x 2 c o s x − 2 ∫ x c o s x d x − 2 c o s x − 2 ∫ x c o s x d x = x 2 c o s x − 2 c o s x − 4 ∫ x c o s x d x = x 2 c o s x − 2 c o s x − 4 x s i n x − 4 c o s x + C = x 2 c o s x − 6 c o s x − 4 x s i n x + C \begin{aligned} &\ \ 设F(x)=\frac{sin\ x}{x},则f(x)=F'(x)=\left(\frac{sin\ x}{x}\right)'=\frac{xcos\ x-sin\ x}{x^2}=\frac{cos\ x}{x}-\frac{sin\ x}{x^2},\\\\ &\ \ 则f'(x)=\frac{-xsin\ x-cos\ x}{x^2}-\frac{x^2cos\ x-2xsin\ x}{x^4}=-\frac{(x^2-2)sin\ x+2xcos\ x}{x^3},\\\\ &\ \ \int x^3f'(x)dx=-\int ((x^2-2)sin\ x+2xcos\ x)dx=-\int x^2sin\ xdx+2\int sin\ xdx-2\int xcos\ xdx=\\\\ &\ \ x^2cos\ x-2\int xcos\ xdx-2cos\ x-2\int xcos\ xdx=x^2cos\ x-2cos\ x-4\int xcos\ xdx=\\\\ &\ \ x^2cos\ x-2cos\ x-4xsin\ x-4cos\ x+C=x^2cos\ x-6cos\ x-4xsin\ x+C & \end{aligned} 设F(x)=xsin x,则f(x)=F′(x)=(xsin x)′=x2xcos x−sin x=xcos x−x2sin x, 则f′(x)=x2−xsin x−cos x−x4x2cos x−2xsin x=−x3(x2−2)sin x+2xcos x, ∫x3f′(x)dx=−∫((x2−2)sin x+2xcos x)dx=−∫x2sin xdx+2∫sin xdx−2∫xcos xdx= x2cos x−2∫xcos xdx−2cos x−2∫xcos xdx=x2cos x−2cos x−4∫xcos xdx= x2cos x−2cos x−4xsin x−4cos x+C=x2cos x−6cos x−4xsin x+C
4. 求下列不定积分(其中 a , b 为常数): \begin{aligned}&4. \ 求下列不定积分(其中a,b为常数):&\end{aligned} 4. 求下列不定积分(其中a,b为常数):
( 1 ) ∫ d x e x − e − x ; ( 2 ) ∫ x ( 1 − x ) 3 d x ; ( 3 ) ∫ x 2 a 6 − x 6 d x ( a > 0 ) ; ( 4 ) ∫ 1 + c o s x x + s i n x d x ; ( 5 ) ∫ l n l n x x d x ; ( 6 ) ∫ s i n x c o s x 1 + s i n 4 x d x ; ( 7 ) ∫ t a n 4 x d x ; ( 8 ) ∫ s i n x s i n 2 x s i n 3 x d x ; ( 9 ) ∫ d x x ( x 6 + 4 ) ; ( 10 ) ∫ a + x a − x d x ( a > 0 ) ; ( 11 ) ∫ d x x ( 1 + x ) ; ( 12 ) ∫ x c o s 2 x d x ; ( 13 ) ∫ e a x c o s b x d x ; ( 14 ) ∫ d x 1 + e x ; ( 15 ) ∫ d x x 2 x 2 − 1 ; ( 16 ) ∫ d x ( a 2 − x 2 ) 5 2 ; ( 17 ) ∫ d x x 4 1 + x 2 ; ( 18 ) ∫ x s i n x d x ; ( 19 ) ∫ l n ( 1 + x 2 ) d x ; ( 20 ) ∫ s i n 2 x c o s 3 x d x ; ( 21 ) ∫ a r c t a n x d x ; ( 22 ) ∫ 1 + c o s x s i n x d x ; ( 23 ) ∫ x 3 ( 1 + x 8 ) 2 d x ; ( 24 ) ∫ x 11 x 8 + 3 x 4 + 2 d x ; ( 25 ) ∫ d x 16 − x 4 ; ( 26 ) ∫ s i n x 1 + s i n x d x ; ( 27 ) ∫ x + s i n x 1 + c o s x d x ; ( 28 ) ∫ e s i n x x c o s 3 x − s i n x c o s 2 x d x ; ( 29 ) ∫ x 3 x ( x + x 3 ) d x ; ( 30 ) ∫ d x ( 1 + e x ) 2 ; ( 31 ) ∫ e 3 x + e x e 4 x − e 2 x + 1 d x ; ( 32 ) ∫ x e x ( e x + 1 ) 2 d x ; ( 33 ) ∫ l n 2 ( x + 1 + x 2 ) d x ; ( 34 ) ∫ l n x ( 1 + x 2 ) 3 2 d x ; ( 35 ) ∫ 1 − x 2 a r c s i n x d x ; ( 36 ) ∫ x 3 a r c c o s x 1 − x 2 d x ; ( 37 ) ∫ c o t x 1 + s i n x d x ; ( 38 ) ∫ d x s i n 3 x c o s x ; ( 39 ) ∫ d x ( 2 + c o s x ) s i n x ; ( 40 ) ∫ s i n x c o s x s i n x + c o s x d x \begin{aligned} &\ \ (1)\ \ \int \frac{dx}{e^x-e^{-x}};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ \int \frac{x}{(1-x)^3}dx;\\\\ &\ \ (3)\ \ \int \frac{x^2}{a^6-x^6}dx\ (a \gt 0);\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ \int \frac{1+cos\ x}{x+sin\ x}dx;\\\\ &\ \ (5)\ \ \int \frac{ln\ ln\ x}{x}dx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)\ \ \int \frac{sin\ xcos\ x}{1+sin^4\ x}dx;\\\\ &\ \ (7)\ \ \int tan^4\ xdx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)\ \ \int sin\ xsin\ 2xsin\ 3xdx;\\\\ &\ \ (9)\ \ \int \frac{dx}{x(x^6+4)};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (10)\ \ \int \sqrt{\frac{a+x}{a-x}}dx\ (a \gt 0);\\\\ &\ \ (11)\ \ \int \frac{dx}{\sqrt{x(1+x)}};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (12)\ \ \int xcos^2\ xdx;\\\\ &\ \ (13)\ \ \int e^{ax}cos\ bxdx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (14)\ \ \int \frac{dx}{\sqrt{1+e^x}};\\\\\ &\ \ (15)\ \ \int \frac{dx}{x^2\sqrt{x^2-1}};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (16)\ \ \int \frac{dx}{(a^2-x^2)^{\frac{5}{2}}};\\\\ &\ \ (17)\ \ \int \frac{dx}{x^4\sqrt{1+x^2}};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (18)\ \ \int \sqrt{x}sin\ \sqrt{x}dx;\\\\ &\ \ (19)\ \ \int ln(1+x^2)dx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (20)\ \ \int \frac{sin^2\ x}{cos^3\ x}dx;\\\\ &\ \ (21)\ \ \int arctan\ \sqrt{x}dx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (22)\ \ \int \frac{\sqrt{1+cos\ x}}{sin\ x}dx;\\\\ &\ \ (23)\ \ \int \frac{x^3}{(1+x^8)^2}dx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (24)\ \ \int \frac{x^{11}}{x^8+3x^4+2}dx;\\\\ &\ \ (25)\ \int \frac{dx}{16-x^4};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (26)\ \ \int \frac{sin\ x}{1+sin\ x}dx;\\\\ &\ \ (27)\ \ \int \frac{x+sin\ x}{1+cos\ x}dx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (28)\ \ \int e^{sin\ x}\frac{xcos^3\ x-sin\ x}{cos^2\ x}dx;\\\\ &\ \ (29)\ \ \int \frac{\sqrt[3]{x}}{x(\sqrt{x}+\sqrt[3]{x})}dx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (30)\ \ \int \frac{dx}{(1+e^x)^2};\\\\ &\ \ (31)\ \int \frac{e^{3x}+e^x}{e^{4x}-e^{2x}+1}dx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (32)\ \ \int \frac{xe^x}{(e^x+1)^2}dx;\\\\ &\ \ (33)\ \ \int ln^2(x+\sqrt{1+x^2})dx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (34)\ \ \int \frac{ln\ x}{(1+x^2)^{\frac{3}{2}}}dx;\\\\ &\ \ (35)\ \ \int \sqrt{1-x^2}arcsin\ xdx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (36)\ \ \int \frac{x^3arccos\ x}{\sqrt{1-x^2}}dx;\\\\ &\ \ (37)\ \ \int \frac{cot\ x}{1+sin\ x}dx;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (38)\ \ \int \frac{dx}{sin^3\ xcos\ x};\\\\ &\ \ (39)\ \ \int \frac{dx}{(2+cos\ x)sin\ x};\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (40)\ \ \int \frac{sin\ xcos\ x}{sin\ x+cos\ x}dx & \end{aligned} (1) ∫ex−e−xdx; (2) ∫(1−x)3xdx; (3) ∫a6−x6x2dx (a>0); (4) ∫x+sin x1+cos xdx; (5) ∫xln ln xdx; (6) ∫1+sin4 xsin xcos xdx; (7) ∫tan4 xdx; (8) ∫sin xsin 2xsin 3xdx; (9) ∫x(x6+4)dx; (10) ∫a−xa+xdx (a>0); (11) ∫x(1+x)dx; (12) ∫xcos2 xdx; (13) ∫eaxcos bxdx; (14) ∫1+exdx; (15) ∫x2x2−1dx; (16) ∫(a2−x2)25dx; (17) ∫x41+x2dx; (18) ∫xsin xdx; (19) ∫ln(1+x2)dx; (20) ∫cos3 xsin2 xdx; (21) ∫arctan xdx; (22) ∫sin x1+cos xdx; (23) ∫(1+x8)2x3dx; (24) ∫x8+3x4+2x11dx; (25) ∫16−x4dx; (26) ∫1+sin xsin xdx; (27) ∫1+cos xx+sin xdx; (28) ∫esin xcos2 xxcos3 x−sin xdx; (29) ∫x(x+3x)3xdx; (30) ∫(1+ex)2dx; (31) ∫e4x−e2x+1e3x+exdx; (32) ∫(ex+1)2xexdx; (33) ∫ln2(x+1+x2)dx; (34) ∫(1+x2)23ln xdx; (35) ∫1−x2arcsin xdx; (36) ∫1−x2x3arccos xdx; (37) ∫1+sin xcot xdx; (38) ∫sin3 xcos xdx; (39) ∫(2+cos x)sin xdx; (40) ∫sin x+cos xsin xcos xdx
解:
( 1 ) ∫ d x e x − e − x = ∫ e x e 2 x − 1 d x ,设 u = e x ,则 x = l n u , d x = 1 u d u ,得 ∫ e x e 2 x − 1 d x = ∫ 1 u 2 − 1 d u = ∫ 1 ( u + 1 ) ( u − 1 ) d u = 2 ∫ 1 u − 1 d u − 2 ∫ 1 u + 1 d u = 2 l n ∣ u − 1 ∣ − 2 l n ∣ u + 1 ∣ + C = 2 l n ∣ e x − 1 ∣ e x + 1 + C ( 2 ) 设 u = 1 − x ,则 x = 1 − u , d x = − d u ,得 ∫ x ( 1 − x ) 3 d x = ∫ u − 1 u 3 d u = ∫ 1 u 2 d u − ∫ 1 u 3 d u = − 1 u + 1 2 u 2 + C = 2 x − 1 2 x 2 − 4 x + 2 + C ( 3 ) ∫ x 2 a 6 − x 6 d x = ∫ x 2 ( a 3 ) 2 − ( x 3 ) 2 d x ,设 u = x 3 ,则 x = x 3 , d x = 1 3 x 2 3 ,得 ∫ x 2 ( a 3 ) 2 − ( x 3 ) 2 d x = ∫ u 2 3 ( a 3 ) 2 − u 2 ⋅ 1 3 x 2 3 d u = − 1 3 ∫ 1 u 2 − ( a 3 ) 2 d u = − 1 3 ( 1 2 a 3 l n ∣ u − a 3 u + a 3 ∣ + C ) = − 1 6 a 3 l n ∣ x 3 − a 3 x 3 + a 3 ∣ + C ( 4 ) ∫ 1 + c o s x x + s i n x d x = ∫ 1 x + s i n x d ( x + s i n x ) = l n ∣ x + s i n x ∣ + C ( 5 ) 设 u = l n x ,则 x = e u , d x = e u d u ,得 ∫ l n l n x x d x = ∫ l n u d u ,设 t = l n u ,则 u = e t , d u = e t d t ,得 ∫ l n u d u = ∫ t e t d t ,设 s = t , d v = e t d t ,则 d s = d t , v = e t ,得 ∫ t e t d t = t e t − e t + C = e t ( t − 1 ) + C = u ( l n u − 1 ) + C = l n x ( l n l n x − 1 ) + C ( 6 ) 设 u = s i n x ,则 x = a r c s i n u , d x = 1 1 − u 2 d u ,得 ∫ s i n x c o s x 1 + s i n 4 x d x = ∫ u 1 − u 2 1 + u 4 ⋅ 1 1 − u 2 d u = ∫ u 1 + u 4 d u = 1 2 ∫ 1 1 + ( u 2 ) 2 d ( u 2 ) = 1 2 a r c t a n ( s i n 2 x ) + C ( 7 ) ∫ t a n 4 x d x = ∫ t a n 2 x ( s e c 2 x − 1 ) d x = ∫ t a n 2 x s e c 2 x d x − ∫ ( s e c 2 x − 1 ) d x = ∫ t a n 2 x d ( t a n x ) − t a n x + x = 1 3 t a n 3 x − t a n x + x + C ( 8 ) ∫ s i n x s i n 2 x s i n 3 x d x = ∫ s i n x ⋅ 2 s i n x c o s x ⋅ ( 3 s i n x − 4 s i n 3 x ) d x = ∫ ( 6 s i n 3 x − 8 s i n 5 x ) c o s x d x , 设 u = s i n x ,则 x = a r c s i n u , d x = 1 1 − u 2 d u ,得 ∫ ( 6 s i n 3 x − 8 s i n 5 x ) c o s x d x = ∫ ( 6 u 3 − 8 u 5 ) d u = 6 ∫ u 3 d u − 8 ∫ u 5 d u = 3 2 u 4 − 4 3 u 6 + C = 3 2 s i n 4 x − 4 3 s i n 6 x + C ( 9 ) 设 u = x 3 ,则 x = x 3 , d x = 1 3 u 2 3 d u ,得 ∫ d x x ( x 6 + 4 ) = 1 3 ∫ 1 u ( u 2 + 4 ) d u ,设 t = u 2 + 4 ,则 u = t − 4 , d u = 1 2 t − 4 d t ,得 1 3 ∫ 1 u ( u 2 + 4 ) d u = 1 6 ∫ 1 t ( t − 4 ) d t = 1 24 ∫ ( 1 t − 4 − 1 t ) d t = 1 24 ∫ 1 t − 4 d ( t − 4 ) − 1 24 ∫ 1 t d t = 1 24 l n ∣ t − 4 ∣ − 1 24 l n ∣ t ∣ + C = 1 4 l n ∣ x ∣ − 1 24 l n ( x 6 + 4 ) + C ( 10 ) ∫ a + x a − x d x = ∫ a + x a 2 − x 2 d x = a ∫ 1 a 2 − x 2 d x + ∫ x a 2 − x 2 d x = a a r c s i n x a − 1 2 ∫ 1 a 2 − x 2 d ( a 2 − x 2 ) = a a r c s i n x a − a 2 − x 2 + C ( 11 ) 设 u = x ( 1 + x ) ,则 x = u 2 + 1 4 − 1 2 , d x = u u 2 + 1 4 d u ,得 ∫ d x x ( 1 + x ) = ∫ 1 u 2 + 1 4 d u = l n ∣ u + u 2 + 1 4 ∣ + C = l n ∣ 2 x ( 1 + x ) + 2 x + 1 ∣ + C ( 12 ) ∫ x c o s 2 x d x = 1 2 ∫ ( x + x c o s 2 x ) d x = 1 2 ∫ x d x + 1 4 ∫ x c o s 2 x d ( 2 x ) = 1 4 x 2 + 1 4 ( x s i n 2 x − 1 2 ∫ s i n 2 x d ( 2 x ) ) = 1 4 x 2 + 1 4 x s i n 2 x + 1 8 c o s 2 x + C ( 13 ) 当 a ≠ 0 时, ∫ e a x c o s b x d x = 1 b ∫ e a x c o s b x d ( b x ) ,设 u = e a x , d v = c o s b x d ( b x ) , 则 d u = a e a x d x , v = s i n b x ,得 1 b ∫ e a x c o s b x d ( b x ) = 1 b e a x s i n b x − a b 2 ∫ e a x s i n b x d ( b x ) ,设 u = e a x , d v = s i n b x d ( b x ) , 则 d u = a e a x d x , v = − c o s b x ,得 1 b e a x s i n b x − a b 2 ∫ e a x s i n b x d ( b x ) = 1 b e a x s i n b x + a b 2 e a x c o s b x − a 2 b 2 ∫ e a x c o s b x d x ,得 ∫ e a x c o s b x d x = e a x a 2 + b 2 ( b s i n b x + a c o s b x ) + C 当 a = 0 时, ∫ e a x c o s b x d x = { s i n b x b + C , b ≠ 0 , x + C , b = 0 ( 14 ) 设 u = 1 + e x ,则 x = l n ( u 2 − 1 ) , d x = 2 u u 2 − 1 d u ,得 ∫ d x 1 + e x = 2 ∫ 1 u 2 − 1 d u = l n ∣ u − a u + a ∣ + C = l n 1 + e x − 1 1 + e x + 1 + C ( 15 ) 令 x = s e c u ,则 x = a r c c o s 1 x , d x = s e c u t a n u d u ,得 ∫ d x x 2 x 2 − 1 = ∫ s e c u t a n u s e c 2 u t a n u d u = ∫ c o s u d u = s i n u + C = s i n ( a r c c o s 1 x ) + C = x 2 − 1 x + C ( 16 ) 设 x = a s i n u ( − π 2 < u < π 2 ) ,则 a 2 − x 2 = a c o s u , d x = a c o s u d u ,得 ∫ d x ( a 2 − x 2 ) 5 2 = 1 a 4 ∫ s e c 4 u d u = 1 a 4 ∫ ( t a n 2 u + 1 ) d ( t a n u ) = 1 3 a 4 t a n 3 u + 1 a 4 t a n u + C = 1 3 a 4 [ x 3 ( a 2 − x 2 ) 3 + 3 x a 2 − x 2 ] + C ( 17 ) 设 x = t a n u ,则 u = a r c t a n x , d x = s e c 2 u d u ,得 ∫ d x x 4 1 + x 2 = ∫ s e c 2 u d u t a n 4 u 1 + t a n 2 u = ∫ s e c u d u t a n 4 u = ∫ c o t 3 u c s c u d u = − ∫ c o t 2 u ( − c s c u c o t u ) d u , 令 s = c o t 2 u , d v = − c s c u c o t u d u ,则 d s = − 2 c o t u c s c 2 u d u , v = c s c u ,得 − ∫ c o t 2 u ( − c s c u c o t u ) d u = − ( c s c u c o t 2 u + 2 ∫ c s c 3 u c o t u d u ) = − ( c s c u c o t 2 u − 2 ∫ c s c 2 u ( − c s c u c o t u ) d u ) = − ( c s c u c o t 2 u − 2 ∫ c s c 2 u d ( c s c u ) ) = 2 3 c s c 3 u − c s c u c o t 2 u + C = 2 3 ( 1 + x 2 ) 3 x 3 − 1 + x 2 x 3 + C ( 18 ) 设 u = x ,则 x = u 2 , d x = 2 u d u ,得 ∫ x s i n x d x = 2 ∫ u 2 s i n u d u ,令 s = u 2 , d v = s i n u d u , d s = 2 u d u , v = − c o s u ,得 2 ∫ u 2 s i n u d u = − 2 u 2 c o s u + 4 ∫ u c o s u d u ,令 s = u , d v = c o s u d u , d s = d u , v = s i n u ,得 − 2 u 2 c o s u + 4 ∫ u c o s u d u = − 2 u 2 c o s u + 4 u s i n u + 4 c o s u + C = − 2 x c o s x + 4 x s i n x + 4 c o s x + C ( 19 ) 设 u = l n ( 1 + x 2 ) ,则 x = e u − 1 , d x = e u d u 2 e u − 1 ,得 ∫ l n ( 1 + x 2 ) d x = 1 2 ∫ u e u d u e u − 1 ,令 s = e u − 1 ,则 u = l n ( s + 1 ) , d u = 1 s + 1 d s ,得 1 2 ∫ u e u d u e u − 1 = 1 2 ∫ l n ( s + 1 ) s d s ,令 u = l n ( s + 1 ) , d v = 1 s d s ,则 d u = 1 s + 1 d s , v = 2 s ,得 1 2 ∫ l n ( s + 1 ) s d s = s l n ( s + 1 ) − ∫ s ( s ) 2 + 1 d s = s l n ( s + 1 ) − 1 2 l n ( s + 1 ) + C = x l n ( x 2 + 1 ) − 1 2 l n ( x 2 + 1 ) + C \begin{aligned} &\ \ (1)\ \int \frac{dx}{e^x-e^{-x}}=\int \frac{e^x}{e^{2x}-1}dx,设u=e^x,则x=ln\ u,dx=\frac{1}{u}du,得\\\\ &\ \ \ \ \ \ \ \ \int \frac{e^x}{e^{2x}-1}dx=\int \frac{1}{u^2-1}du=\int \frac{1}{(u+1)(u-1)}du=2\int \frac{1}{u-1}du-2\int \frac{1}{u+1}du=\\\\ &\ \ \ \ \ \ \ \ 2ln\ |u-1|-2ln\ |u+1|+C=2ln\ \frac{|e^x-1|}{e^x+1}+C\\\\ &\ \ (2)\ 设u=1-x,则x=1-u,dx=-du,得\\\\ &\ \ \ \ \ \ \ \ \int \frac{x}{(1-x)^3}dx=\int \frac{u-1}{u^3}du=\int \frac{1}{u^2}du-\int \frac{1}{u^3}du=-\frac{1}{u}+\frac{1}{2u^2}+C=\frac{2x-1}{2x^2-4x+2}+C\\\\ &\ \ (3)\ \int \frac{x^2}{a^6-x^6}dx=\int \frac{x^2}{(a^3)^2-(x^3)^2}dx,设u=x^3,则x=\sqrt[3]{x},dx=\frac{1}{3\sqrt[3]{x^2}},得\\\\ &\ \ \ \ \ \ \ \ \int \frac{x^2}{(a^3)^2-(x^3)^2}dx=\int \frac{\sqrt[3]{u^2}}{(a^3)^2-u^2}\cdot \frac{1}{3\sqrt[3]{x^2}}du=-\frac{1}{3}\int \frac{1}{u^2-(a^3)^2}du=-\frac{1}{3}\left(\frac{1}{2a^3}ln\ \left|\frac{u-a^3}{u+a^3}\right|+C\right)=\\\\ &\ \ \ \ \ \ \ \ \ \ -\frac{1}{6a^3}ln\ \left|\frac{x^3-a^3}{x^3+a^3}\right|+C\\\\ &\ \ (4)\ \int \frac{1+cos\ x}{x+sin\ x}dx=\int \frac{1}{x+sin\ x}d(x+sin\ x)=ln\ |x+sin\ x|+C\\\\ &\ \ (5)\ 设u=ln\ x,则x=e^u,dx=e^udu,得\int \frac{ln\ ln\ x}{x}dx=\int ln\ udu,设t=ln\ u,则u=e^t,du=e^tdt,得\\\\ &\ \ \ \ \ \ \ \ \int ln\ udu=\int te^tdt,设s=t,dv=e^tdt,则ds=dt,v=e^t,得\\\\ &\ \ \ \ \ \ \ \ \int te^tdt=te^t-e^t+C=e^t(t-1)+C=u(ln\ u-1)+C=ln\ x(ln\ ln\ x-1)+C\\\\ &\ \ (6)\ 设u=sin\ x,则x=arcsin\ u,dx=\frac{1}{\sqrt{1-u^2}}du,得\\\\ &\ \ \ \ \ \ \ \ \int \frac{sin\ xcos\ x}{1+sin^4\ x}dx=\int \frac{u\sqrt{1-u^2}}{1+u^4}\cdot \frac{1}{\sqrt{1-u^2}}du=\int \frac{u}{1+u^4}du=\frac{1}{2}\int \frac{1}{1+(u^2)^2}d(u^2)=\frac{1}{2}arctan(sin^2\ x)+C\\\\ &\ \ (7)\ \int tan^4\ xdx=\int tan^2\ x(sec^2\ x-1)dx=\int tan^2\ xsec^2\ xdx-\int (sec^2\ x-1)dx=\\\\ &\ \ \ \ \ \ \ \ \int tan^2\ xd(tan\ x)-tan\ x+x=\frac{1}{3}tan^3\ x-tan\ x+x+C\\\\ &\ \ (8)\ \int sin\ xsin\ 2xsin\ 3xdx=\int sin\ x\cdot 2sin\ xcos\ x\cdot (3sin\ x-4sin^3\ x)dx=\int (6sin^3\ x-8sin^5\ x)cos\ xdx,\\\\ &\ \ \ \ \ \ \ \ 设u=sin\ x,则x=arcsin\ u,dx=\frac{1}{\sqrt{1-u^2}}du,得\\\\ &\ \ \ \ \ \ \ \ \int (6sin^3\ x-8sin^5\ x)cos\ xdx=\int (6u^3-8u^5)du=6\int u^3du-8\int u^5du=\\\\ &\ \ \ \ \ \ \ \ \frac{3}{2}u^4-\frac{4}{3}u^6+C=\frac{3}{2}sin^4\ x-\frac{4}{3}sin^6\ x+C\\\\ &\ \ (9)\ 设u=x^3,则x=\sqrt[3]{x},dx=\frac{1}{3\sqrt[3]{u^2}}du,得\\\\ &\ \ \ \ \ \ \ \ \int \frac{dx}{x(x^6+4)}=\frac{1}{3}\int \frac{1}{u(u^2+4)}du,设t=u^2+4,则u=\sqrt{t-4},du=\frac{1}{2\sqrt{t-4}}dt,得\\\\ &\ \ \ \ \ \ \ \ \frac{1}{3}\int \frac{1}{u(u^2+4)}du=\frac{1}{6}\int \frac{1}{t(t-4)}dt=\frac{1}{24}\int \left(\frac{1}{t-4}-\frac{1}{t}\right)dt=\frac{1}{24}\int \frac{1}{t-4}d(t-4)-\frac{1}{24}\int \frac{1}{t}dt=\\\\ &\ \ \ \ \ \ \ \ \frac{1}{24}ln\ |t-4|-\frac{1}{24}ln\ |t|+C=\frac{1}{4}ln\ |x|-\frac{1}{24}ln(x^6+4)+C\\\\ &\ \ (10)\ \int \sqrt{\frac{a+x}{a-x}}dx=\int \frac{a+x}{\sqrt{a^2-x2}}dx=a\int \frac{1}{\sqrt{a^2-x^2}}dx+\int \frac{x}{\sqrt{a^2-x^2}}dx=\\\\ &\ \ \ \ \ \ \ \ \ \ aarcsin\ \frac{x}{a}-\frac{1}{2}\int \frac{1}{\sqrt{a^2-x^2}}d(a^2-x^2)=aarcsin\ \frac{x}{a}-\sqrt{a^2-x^2}+C\\\\ &\ \ (11)\ 设u=\sqrt{x(1+x)},则x=\sqrt{u^2+\frac{1}{4}}-\frac{1}{2},dx=\frac{u}{\sqrt{u^2+\frac{1}{4}}}du,得\\\\ &\ \ \ \ \ \ \ \ \ \int \frac{dx}{\sqrt{x(1+x)}}=\int \frac{1}{\sqrt{u^2+\frac{1}{4}}}du=ln\ |u+\sqrt{u^2+\frac{1}{4}}|+C=ln\ |2\sqrt{x(1+x)}+2x+1|+C\\\\ &\ \ (12)\ \int xcos^2\ xdx=\frac{1}{2}\int (x+xcos\ 2x)dx=\frac{1}{2}\int xdx+\frac{1}{4}\int xcos\ 2xd(2x)=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{4}x^2+\frac{1}{4}\left(xsin\ 2x-\frac{1}{2}\int sin\ 2xd(2x)\right)=\frac{1}{4}x^2+\frac{1}{4}xsin\ 2x+\frac{1}{8}cos\ 2x+C\\\\ &\ \ (13)\ 当a \neq 0时,\int e^{ax}cos\ bxdx=\frac{1}{b}\int e^{ax}cos\ bxd(bx),设u=e^{ax},dv=cos\ bxd(bx),\\\\ &\ \ \ \ \ \ \ \ \ \ \ 则du=ae^{ax}dx,v=sin\ bx,得\\\\ &\ \ \ \ \ \ \ \ \ \ \ \frac{1}{b}\int e^{ax}cos\ bxd(bx)=\frac{1}{b}e^{ax}sin\ bx-\frac{a}{b^2}\int e^{ax}sin\ bxd(bx),设u=e^{ax},dv=sin\ bxd(bx),\\\\ &\ \ \ \ \ \ \ \ \ \ \ 则du=ae^{ax}dx,v=-cos\ bx,得\\\\ &\ \ \ \ \ \ \ \ \ \ \ \frac{1}{b}e^{ax}sin\ bx-\frac{a}{b^2}\int e^{ax}sin\ bxd(bx)=\frac{1}{b}e^{ax}sin\ bx+\frac{a}{b^2}e^{ax}cos\ bx-\frac{a^2}{b^2}\int e^{ax}cos\ bxdx,得\\\\ &\ \ \ \ \ \ \ \ \ \ \ \int e^{ax}cos\ bxdx=\frac{e^{ax}}{a^2+b^2}(bsin\ bx+acos\ bx)+C\\\\ &\ \ \ \ \ \ \ \ \ \ \ 当a=0时,\int e^{ax}cos\ bxdx=\begin{cases}\frac{sin\ bx}{b}+C,b \neq 0,\\\\x+C,b=0\end{cases}\\\\ &\ \ (14)\ 设u=\sqrt{1+e^x},则x=ln(u^2-1),dx=\frac{2u}{u^2-1}du,得\\\\ &\ \ \ \ \ \ \ \ \ \ \ \int \frac{dx}{\sqrt{1+e^x}}=2\int \frac{1}{u^2-1}du=ln\ \left|\frac{u-a}{u+a}\right|+C=ln\ \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}+C\\\\ &\ \ (15)\ 令x=sec\ u,则x=arccos\ \frac{1}{x},dx=sec\ utan\ udu,得\\\\ &\ \ \ \ \ \ \ \ \ \ \ \int \frac{dx}{x^2\sqrt{x^2-1}}=\int \frac{sec\ utan\ u}{sec^2\ utan\ u}du=\int cos\ udu=sin\ u+C=sin(arccos\ \frac{1}{x})+C=\frac{\sqrt{x^2-1}}{x}+C\\\\ &\ \ (16)\ 设x=asin\ u\ \left(-\frac{\pi}{2} \lt u \lt \frac{\pi}{2}\right),则\sqrt{a^2-x^2}=acos\ u,dx=acos\ udu,得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{dx}{(a^2-x^2)^{\frac{5}{2}}}=\frac{1}{a^4}\int sec^4\ udu=\frac{1}{a^4}\int (tan^2\ u+1)d(tan\ u)=\frac{1}{3a^4}tan^3\ u+\frac{1}{a^4}tan\ u+C=\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{3a^4}\left[\frac{x^3}{\sqrt{(a^2-x^2)^3}}+\frac{3x}{\sqrt{a^2-x^2}}\right]+C\\\\ &\ \ (17)\ 设x=tan\ u,则u=arctan\ x,dx=sec^2\ udu,得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \frac{dx}{x^4\sqrt{1+x^2}}=\int \frac{sec^2\ udu}{tan^4\ u\sqrt{1+tan^2\ u}}=\int \frac{sec\ udu}{tan^4\ u}=\int cot^3\ ucsc\ udu=-\int cot^2\ u(-csc\ ucot\ u)du,\\\\ &\ \ \ \ \ \ \ \ \ \ 令s=cot^2\ u,dv=-csc\ ucot\ udu,则ds=-2cot\ ucsc^2\ udu,v=csc\ u,得\\\\ &\ \ \ \ \ \ \ \ \ \ -\int cot^2\ u(-csc\ ucot\ u)du=-(csc\ ucot^2\ u+2\int csc^3\ ucot\ udu)=\\\\ &\ \ \ \ \ \ \ \ \ \ -(csc\ ucot^2\ u-2\int csc^2\ u(-csc\ ucot\ u)du)=-(csc\ ucot^2\ u-2\int csc^2\ ud(csc\ u))=\\\\ &\ \ \ \ \ \ \ \ \ \ \ \frac{2}{3}csc^3\ u-csc\ ucot^2\ u+C=\frac{2}{3}\frac{\sqrt{(1+x^2)^3}}{x^3}-\frac{\sqrt{1+x^2}}{x^3}+C\\\\ &\ \ (18)\ 设u=\sqrt{x},则x=u^2,dx=2udu,得\\\\ &\ \ \ \ \ \ \ \ \ \ \int \sqrt{x}sin\ \sqrt{x}dx=2\int u^2sin\ udu,令s=u^2,dv=sin\ udu,ds=2udu,v=-cos\ u,得\\\\ &\ \ \ \ \ \ \ \ \ \ 2\int u^2sin\ udu=-2u^2cos\ u+4\int ucos\ udu,令s=u,dv=cos\ udu,ds=du,v=sin\ u,得\\\\ &\ \ \ \ \ \ \ \ \ \ -2u^2cos\ u+4\int ucos\ udu=-2u^2cos\ u+4usin\ u+4cos\ u+C=-2xcos\sqrt{x}+4\sqrt{x}sin\sqrt{x}+4cos\sqrt{x}+C\\\\ &\ \ (19)\ 设u=ln(1+x^2),则x=\sqrt{e^u-1},dx=\frac{e^udu}{2\sqrt{e^u-1}},得\\\\ &\ \ \ \ \ \ \ \ \ \ \int ln(1+x^2)dx=\frac{1}{2}\int \frac{ue^udu}{\sqrt{e^u-1}},令s=e^u-1,则u=ln(s+1),du=\frac{1}{s+1}ds,得\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}\int \frac{ue^udu}{\sqrt{e^u-1}}=\frac{1}{2}\int \frac{ln(s+1)}{\sqrt{s}}ds,令u=ln(s+1),dv=\frac{1}{\sqrt{s}}ds,则du=\frac{1}{s+1}ds,v=2\sqrt{s},得\\\\ &\ \ \ \ \ \ \ \ \ \ \frac{1}{2}\int \frac{ln(s+1)}{\sqrt{s}}ds=\sqrt{s}ln(s+1)-\int \frac{\sqrt{s}}{(\sqrt{s})^2+1}ds=\sqrt{s}ln(s+1)-\frac{1}{2}ln(s+1)+C=\\\\ &\ \ \ \ \ \ \ \ \ \ xln(x^2+1)-\frac{1}{2}ln(x^2+1)+C & \end{aligned} (1) ∫ex−e−xdx=∫e2x−1exdx,设u=ex,则x=ln u,dx=u1du,得 ∫e2x−1exdx=∫u2−11du=∫(u+1)(u−1)1du=2∫u−11du−2∫u+11du= 2ln ∣u−1∣−2ln ∣u+1∣+C=2ln ex+1∣ex−1∣+C (2) 设u=1−x,则x=1−u,dx=−du,得 ∫(1−x)3xdx=∫u3u−1du=∫u21du−∫u31du=−u1+2u21+C=2x2−4x+22x−1+C (3) ∫a6−x6x2dx=∫(a3)2−(x3)2x2dx,设u=x3,则x=3x,dx=33x21,得 ∫(a3)2−(x3)2x2dx=∫(a3)2−u23u2⋅33x21du=−31∫u2−(a3)21du=−31(2a31ln ∣ ∣u+a3u−a3∣ ∣+C)= −6a31ln ∣ ∣x3+a3x3−a3∣ ∣+C (4) ∫x+sin x1+cos xdx=∫x+sin x1d(x+sin x)=ln ∣x+sin x∣+C (5) 设u=ln x,则x=eu,dx=eudu,得∫xln ln xdx=∫ln udu,设t=ln u,则u=et,du=etdt,得 ∫ln udu=∫tetdt,设s=t,dv=etdt,则ds=dt,v=et,得 ∫tetdt=tet−et+C=et(t−1)+C=u(ln u−1)+C=ln x(ln ln x−1)+C (6) 设u=sin x,则x=arcsin u,dx=1−u21du,得 ∫1+sin4 xsin xcos xdx=∫1+u4u1−u2⋅1−u21du=∫1+u4udu=21∫1+(u2)21d(u2)=21arctan(sin2 x)+C (7) ∫tan4 xdx=∫tan2 x(sec2 x−1)dx=∫tan2 xsec2 xdx−∫(sec2 x−1)dx= ∫tan2 xd(tan x)−tan x+x=31tan3 x−tan x+x+C (8) ∫sin xsin 2xsin 3xdx=∫sin x⋅2sin xcos x⋅(3sin x−4sin3 x)dx=∫(6sin3 x−8sin5 x)cos xdx, 设u=sin x,则x=arcsin u,dx=1−u21du,得 ∫(6sin3 x−8sin5 x)cos xdx=∫(6u3−8u5)du=6∫u3du−8∫u5du= 23u4−34u6+C=23sin4 x−34sin6 x+C (9) 设u=x3,则x=3x,dx=33u21du,得 ∫x(x6+4)dx=31∫u(u2+4)1du,设t=u2+4,则u=t−4,du=2t−41dt,得 31∫u(u2+4)1du=61∫t(t−4)1dt=241∫(t−41−t1)dt=241∫t−41d(t−4)−241∫t1dt= 241ln ∣t−4∣−241ln ∣t∣+C=41ln ∣x∣−241ln(x6+4)+C (10) ∫a−xa+xdx=∫a2−x2a+xdx=a∫a2−x21dx+∫a2−x2xdx= aarcsin ax−21∫a2−x21d(a2−x2)=aarcsin ax−a2−x2+C (11) 设u=x(1+x),则x=u2+41−21,dx=u2+41udu,得 ∫x(1+x)dx=∫u2+411du=ln ∣u+u2+41∣+C=ln ∣2x(1+x)+2x+1∣+C (12) ∫xcos2 xdx=21∫(x+xcos 2x)dx=21∫xdx+41∫xcos 2xd(2x)= 41x2+41(xsin 2x−21∫sin 2xd(2x))=41x2+41xsin 2x+81cos 2x+C (13) 当a=0时,∫eaxcos bxdx=b1∫eaxcos bxd(bx),设u=eax,dv=cos bxd(bx), 则du=aeaxdx,v=sin bx,得 b1∫eaxcos bxd(bx)=b1eaxsin bx−b2a∫eaxsin bxd(bx),设u=eax,dv=sin bxd(bx), 则du=aeaxdx,v=−cos bx,得 b1eaxsin bx−b2a∫eaxsin bxd(bx)=b1eaxsin bx+b2aeaxcos bx−b2a2∫eaxcos bxdx,得 ∫eaxcos bxdx=a2+b2eax(bsin bx+acos bx)+C 当a=0时,∫eaxcos bxdx=⎩ ⎨ ⎧bsin bx+C,b=0,x+C,b=0 (14) 设u=1+ex,则x=ln(u2−1),dx=u2−12udu,得 ∫1+exdx=2∫u2−11du=ln ∣ ∣u+au−a∣ ∣+C=ln 1+ex+11+ex−1+C (15) 令x=sec u,则x=arccos x1,dx=sec utan udu,得 ∫x2x2−1dx=∫sec2 utan usec utan udu=∫cos udu=sin u+C=sin(arccos x1)+C=xx2−1+C (16) 设x=asin u (−2π<u<2π),则a2−x2=acos u,dx=acos udu,得 ∫(a2−x2)25dx=a41∫sec4 udu=a41∫(tan2 u+1)d(tan u)=3a41tan3 u+a41tan u+C= 3a41[(a2−x2)3x3+a2−x23x]+C (17) 设x=tan u,则u=arctan x,dx=sec2 udu,得 ∫x41+x2dx=∫tan4 u1+tan2 usec2 udu=∫tan4 usec udu=∫cot3 ucsc udu=−∫cot2 u(−csc ucot u)du, 令s=cot2 u,dv=−csc ucot udu,则ds=−2cot ucsc2 udu,v=csc u,得 −∫cot2 u(−csc ucot u)du=−(csc ucot2 u+2∫csc3 ucot udu)= −(csc ucot2 u−2∫csc2 u(−csc ucot u)du)=−(csc ucot2 u−2∫csc2 ud(csc u))= 32csc3 u−csc ucot2 u+C=32x3(1+x2)3−x31+x2+C (18) 设u=x,则x=u2,dx=2udu,得 ∫xsin xdx=2∫u2sin udu,令s=u2,dv=sin udu,ds=2udu,v=−cos u,得 2∫u2sin udu=−2u2cos u+4∫ucos udu,令s=u,dv=cos udu,ds=du,v=sin u,得 −2u2cos u+4∫ucos udu=−2u2cos u+4usin u+4cos u+C=−2xcosx+4xsinx+4cosx+C (19) 设u=ln(1+x2),则x=eu−1,dx=2eu−1eudu,得 ∫ln(1+x2)dx=21∫eu−1ueudu,令s=eu−1,则u=ln(s+1),du=s+11ds,得 21∫eu−1ueudu=21∫sln(s+1)ds,令u=ln(s+1),dv=s1ds,则du=s+11ds,v=2s,得 21∫sln(s+1)ds=sln(s+1)−∫(s)2+1sds=sln(s+1)−21ln(s+1)+C= xln(x2+1)−21ln(x2+1)+C
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