一. 不定积分的概念
1.1 原函数
(
x
2
)
′
=
?
(x^2)'=?
(x2)′=?
\quad
\quad
(
?
)
′
=
2
x
(?)'=2x
(?)′=2x
(
原函数
)
′
=
导数
(原函数)'=导数
(原函数)′=导数
\quad
思考一下
(
?
)
′
=
2
x
(?)'=2x
(?)′=2x
有
x
2
,
x
2
+
1
,
x
2
−
6
,
x
2
−
e
.
.
.
x^2, x^2+1, x^2-6, x^2-e...
x2,x2+1,x2−6,x2−e...
我们用
x
2
+
C
x^2+C
x2+C来代表全体原函数
\quad
1.2 不定积分
F ′ ( x ) = f ( x ) F'(x)=f(x) F′(x)=f(x)
∫
f
(
x
)
d
x
=
F
(
x
)
+
C
\int f(x)dx=F(x)+C
∫f(x)dx=F(x)+C
\quad
(不定积分)
\quad
例题1:
∫
sin
x
d
x
\int \sin x dx
∫sinxdx
(
−
cos
x
)
′
=
sin
x
(-\cos x)'=\sin x
(−cosx)′=sinx
∫
sin
x
d
x
=
−
cos
x
+
C
\int \sin x dx=-\cos x+C
∫sinxdx=−cosx+C
\quad
例题2:
∫
e
x
d
x
=
e
x
+
C
\int e^xdx=e^x+C
∫exdx=ex+C
\quad
例题3:
∫
d
x
=
∫
1
d
x
=
x
+
C
\int dx=\int 1dx=x+C
∫dx=∫1dx=x+C
\quad
例题4:
∫
0
d
x
=
C
\int 0dx=C
∫0dx=C
\quad
例题5:
∫
x
x
d
x
=
x
3
4
+
C
\int \sqrt{x\sqrt{x}}dx=x^{\frac{3}{4}}+C
∫xxdx=x43+C
\quad
\quad
二. 直接积分法
2.1 法则
加减: ∫ ( sin x ± e x ) d x = ∫ sin x d x ± ∫ e x d x \int (\sin x \pm e^x)dx=\int \sin xdx \pm \int e^xdx ∫(sinx±ex)dx=∫sinxdx±∫exdx
数乘: ∫ c sin x d x = c ∫ sin x d x \int c \sin xdx=c\int \sin xdx ∫csinxdx=c∫sinxdx \quad (常数可以提出来)
特别注意: 积分不能分开乘除
如:
∫
sin
x
∗
cos
x
d
x
≠
∫
sin
x
d
x
∗
∫
cos
x
d
x
\int \sin x*\cos xdx \neq \int \sin xdx * \int \cos xdx
∫sinx∗cosxdx=∫sinxdx∗∫cosxdx
∫
x
cos
x
d
x
≠
x
∫
cos
x
d
x
\int x \cos xdx \neq x\int \cos xdx
∫xcosxdx=x∫cosxdx
\quad
(x不能提出来)
∫
cos
x
x
d
x
≠
∫
∫
cos
x
d
x
∫
x
d
x
\int \frac{\cos x}{x}dx \neq \int \frac{\int \cos xdx}{\int x dx}
∫xcosxdx=∫∫xdx∫cosxdx
\quad
2.2 公式
∫ x 3 d x = 1 4 x 4 + C \int x^3 dx=\frac{1}{4}x^4+C ∫x3dx=41x4+C
∫ 3 x d x = 3 x ln 3 + C \int 3^x dx=\frac{3^x}{\ln 3}+C ∫3xdx=ln33x+C
∫ 1 x d x = ln ∣ x ∣ + C \int \frac{1}{x}dx=\ln|x|+C ∫x1dx=ln∣x∣+C
∫ sec 2 x d x = tan x + C \int \sec^2xdx=\tan x+C ∫sec2xdx=tanx+C
∫ c s c 2 x d x = − cot x + C \int csc^2xdx=-\cot x+C ∫csc2xdx=−cotx+C
∫ tan x sec x d x = sec x + C \int \tan x \sec xdx=\sec x+C ∫tanxsecxdx=secx+C
∫ cot x csc d x = − csc x + C \int \cot x \csc dx=-\csc x+C ∫cotxcscdx=−cscx+C
∫ 1 1 − x 2 d x = arcsin x + C \int \frac{1}{\sqrt{1-x^2}}dx=\arcsin x+C ∫1−x21dx=arcsinx+C
∫ − 1 1 − x 2 d x = arccos x + C \int \frac{-1}{\sqrt{1-x^2}}dx=\arccos x+C ∫1−x2−1dx=arccosx+C
∫ 1 1 + x 2 d x = arctan x + C \int \frac{1}{1+x^2}dx=\arctan x+C ∫1+x21dx=arctanx+C
\quad
例题6:
∫
(
x
+
1
)
2
x
d
x
\int \frac{(x+1)^2}{x}dx
∫x(x+1)2dx
=
∫
x
2
+
2
x
+
1
x
d
x
\int \frac{x^2+2x+1}{x}dx
∫xx2+2x+1dx
=
∫
x
+
2
+
1
x
d
x
\int x+2+\frac{1}{x}dx
∫x+2+x1dx
=
1
2
x
2
+
2
x
+
ln
∣
x
∣
+
C
\frac{1}{2}x^2+2x+\ln |x|+C
21x2+2x+ln∣x∣+C
\quad
例题7:
∫
x
2
+
x
e
x
+
1
x
d
x
\int \frac{x^2+\sqrt{x}e^x+1}{\sqrt{x}}dx
∫xx2+xex+1dx
=
∫
x
2
x
+
e
x
+
1
x
d
x
\int \frac{x^2}{\sqrt{x}}+e^x+\frac{1}{\sqrt{x}}dx
∫xx2+ex+x1dx
=
∫
x
x
+
e
x
+
1
x
d
x
\int x\sqrt{x}+e^x+\frac{1}{\sqrt{x}}dx
∫xx+ex+x1dx
=
∫
x
3
2
+
e
x
+
x
−
1
2
d
x
\int x^{\frac{3}{2}}+e^x+x^{-\frac{1}{2}}dx
∫x23+ex+x−21dx
=
2
5
x
5
2
+
e
x
+
2
x
1
2
+
C
\frac{2}{5}x^{\frac{5}{2}}+e^x+2x^{\frac{1}{2}}+C
52x25+ex+2x21+C
\quad
例题8:
∫
3
x
e
x
d
x
\int 3^xe^x dx
∫3xexdx
=
∫
(
3
e
)
x
d
x
\int (3e)^xdx
∫(3e)xdx
=
(
3
e
)
x
ln
3
e
+
C
\frac{(3e)^x}{\ln3e}+C
ln3e(3e)x+C
\quad
例题9:
∫
(
x
+
1
)
2
x
(
1
+
x
2
)
d
x
\int \frac{(x+1)^2}{x(1+x^2)}dx
∫x(1+x2)(x+1)2dx
=
∫
x
2
+
1
x
(
1
+
x
2
)
+
2
x
x
(
1
+
x
2
)
d
x
\int \frac{x^2+1}{x(1+x^2)}+\frac{2x}{x(1+x^2)}dx
∫x(1+x2)x2+1+x(1+x2)2xdx
\quad
例题10:
∫
1
+
3
x
2
x
2
(
1
+
x
2
)
d
x
\int \frac{1+3x^2}{x^2(1+x^2)}dx
∫x2(1+x2)1+3x2dx
=
∫
3
x
2
x
2
(
1
+
x
2
)
+
1
x
2
(
1
+
x
2
)
d
x
\int \frac{3x^2}{x^2(1+x^2)}+\frac{1}{x^2(1+x^2)}dx
∫x2(1+x2)3x2+x2(1+x2)1dx
=
∫
3
(
1
+
x
2
)
+
1
+
x
2
−
x
2
x
2
(
1
+
x
2
)
d
x
\int \frac{3}{(1+x^2)}+\frac{1+x^2-x^2}{x^2(1+x^2)}dx
∫(1+x2)3+x2(1+x2)1+x2−x2dx
=
∫
3
(
1
+
x
2
)
+
1
+
x
2
x
2
(
1
+
x
2
)
−
x
2
x
2
(
1
+
x
2
)
d
x
\int \frac{3}{(1+x^2)}+\frac{1+x^2}{x^2(1+x^2)}-\frac{x^2}{x^2(1+x^2)}dx
∫(1+x2)3+x2(1+x2)1+x2−x2(1+x2)x2dx
=
3
arctan
x
−
x
−
1
−
arctan
x
+
C
3\arctan x-x^{-1}-\arctan x+C
3arctanx−x−1−arctanx+C
=
2
arctan
x
−
x
−
1
+
C
2\arctan x-x^{-1}+C
2arctanx−x−1+C
\quad
例题11:
f
(
x
)
f(x)
f(x)的一个原函数是
ln
x
\ln x
lnx,求
f
′
(
x
)
f'(x)
f′(x)
依题意得:
(
ln
x
)
′
=
f
(
x
)
(\ln x)'=f(x)
(lnx)′=f(x)
∴
\therefore
∴
f
(
x
)
=
1
x
f(x)=\frac{1}{x}
f(x)=x1
∴
\therefore
∴
f
′
(
x
)
=
−
1
x
2
f'(x)=-\frac{1}{x^2}
f′(x)=−x21
\quad
\quad
三. 第一类换元法(凑微分)
推导:
∫
cos
x
d
x
=
sin
x
+
C
\int \cos xdx=\sin x+C
∫cosxdx=sinx+C
->
∫
cos
u
d
u
=
sin
u
+
C
\int \cos udu=\sin u+C
∫cosudu=sinu+C
->
∫
cos
3
x
d
3
x
=
sin
3
x
+
C
\int \cos 3xd3x=\sin3x+C
∫cos3xd3x=sin3x+C
\quad
\quad
例题12:
∫
e
−
x
d
x
\int e^{-x}dx
∫e−xdx
=
−
∫
e
−
x
d
(
−
x
)
-\int e^{-x}d(-x)
−∫e−xd(−x)
=
−
e
−
x
+
C
-e^{-x}+C
−e−x+C
\quad
例题13:
∫
cos
(
3
x
+
2
)
d
x
\int \cos(3x+2)dx
∫cos(3x+2)dx
=
1
3
∫
cos
(
3
x
+
2
)
d
3
x
\frac{1}{3}\int \cos(3x+2)d3x
31∫cos(3x+2)d3x
=
1
3
∫
cos
(
3
x
+
2
)
d
(
3
x
+
2
)
\frac{1}{3}\int \cos(3x+2)d(3x+2)
31∫cos(3x+2)d(3x+2)
\quad
\quad
(后面可以加减常数,结果不变)
=
1
3
sin
(
3
x
+
2
)
+
C
\frac{1}{3} \sin(3x+2)+C
31sin(3x+2)+C
\quad
例题14:
∫
e
1
−
x
d
x
\int e^{1-x}dx
∫e1−xdx
=
−
∫
e
1
−
x
d
(
−
x
)
-\int e^{1-x}d(-x)
−∫e1−xd(−x)
=
−
∫
e
1
−
x
d
(
−
x
+
1
)
-\int e^{1-x}d(-x+1)
−∫e1−xd(−x+1)
=
−
e
1
−
x
+
C
- e^{1-x}+C
−e1−x+C
\quad
例题15:
∫
sin
(
2
x
−
1
)
d
x
\int \sin(2x-1)dx
∫sin(2x−1)dx
=
1
2
∫
sin
(
2
x
−
1
)
d
2
x
\frac{1}{2}\int \sin(2x-1)d2x
21∫sin(2x−1)d2x
=
1
2
∫
sin
(
2
x
−
1
)
d
(
2
x
−
1
)
\frac{1}{2}\int \sin(2x-1)d(2x-1)
21∫sin(2x−1)d(2x−1)
=
−
1
2
cos
(
2
x
−
1
)
+
C
-\frac{1}{2}\cos(2x-1)+C
−21cos(2x−1)+C
\quad
例题16:
∫
2
x
+
1
d
x
\int \sqrt{2x+1}dx
∫2x+1dx
=
1
2
∫
2
x
+
1
d
2
x
\frac{1}{2}\int\sqrt{2x+1}d2x
21∫2x+1d2x
=
1
2
∫
2
x
+
1
d
(
2
x
+
1
)
\frac{1}{2}\int\sqrt{2x+1}d(2x+1)
21∫2x+1d(2x+1)
=
1
2
∗
2
3
(
2
x
+
1
)
3
2
+
C
\frac{1}{2}*\frac{2}{3}(2x+1)^{\frac{3}{2}}+C
21∗32(2x+1)23+C
=
1
3
(
2
x
+
1
)
3
2
+
C
\frac{1}{3}(2x+1)^{\frac{3}{2}}+C
31(2x+1)23+C
\quad
例题17:
∫
1
1
+
4
x
2
d
x
\int \frac{1}{1+4x^2}dx
∫1+4x21dx
=
∫
1
1
+
(
2
x
)
2
d
x
\int \frac{1}{1+(2x)^2}dx
∫1+(2x)21dx
=
1
2
∫
1
1
+
(
2
x
)
2
d
(
2
x
)
\frac{1}{2}\int \frac{1}{1+(2x)^2}d(2x)
21∫1+(2x)21d(2x)
=
1
2
arctan
2
x
+
C
\frac{1}{2}\arctan2x+C
21arctan2x+C
\quad
例题18:
∫
1
2
x
+
1
d
x
\int \frac{1}{2x+1}dx
∫2x+11dx
=
1
2
∫
1
2
x
+
1
d
2
x
\frac{1}{2}\int \frac{1}{2x+1}d2x
21∫2x+11d2x
=
1
2
∫
1
2
x
+
1
d
(
2
x
+
1
)
\frac{1}{2}\int \frac{1}{2x+1}d(2x+1)
21∫2x+11d(2x+1)
=
1
2
ln
(
2
x
+
1
)
+
C
\frac{1}{2}\ln(2x+1)+C
21ln(2x+1)+C
\quad
\quad
例题19:
∫
x
e
x
2
d
x
\int xe^{x^2}dx
∫xex2dx
=
1
2
∫
e
x
2
2
x
d
x
\frac{1}{2}\int e^{x^2}2xdx
21∫ex22xdx
=
1
2
∫
e
x
2
(
x
2
)
′
d
x
\frac{1}{2}\int e^{x^2}(x^2)'dx
21∫ex2(x2)′dx
=
1
2
∫
e
x
2
d
x
2
\frac{1}{2}\int e^{x^2}dx^2
21∫ex2dx2
=
1
2
e
x
2
+
C
\frac{1}{2}e^{x^2}+C
21ex2+C
\quad
例题20:
∫
e
x
x
d
x
\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx
∫xexdx
=
∫
e
x
∗
1
x
d
x
\int e^{\sqrt{x}}*\frac{1}{\sqrt{x}}dx
∫ex∗x1dx
=
∫
e
x
∗
(
2
x
)
′
d
x
\int e^{\sqrt{x}}*(2\sqrt{x})'dx
∫ex∗(2x)′dx
=
∫
e
x
d
(
2
x
)
\int e^{\sqrt{x}}d(2\sqrt{x})
∫exd(2x)
=
2
∫
e
x
d
x
2\int e^{\sqrt{x}}d\sqrt{x}
2∫exdx
=
2
e
x
+
C
2e^{\sqrt{x}}+C
2ex+C
\quad
例题21:
∫
ln
x
x
d
x
\int \frac{\ln x}{x}dx
∫xlnxdx
=
∫
ln
x
∗
1
x
d
x
\int \ln x*\frac{1}{x}dx
∫lnx∗x1dx
=
∫
ln
x
∗
(
ln
x
)
′
d
x
\int \ln x*(\ln x)'dx
∫lnx∗(lnx)′dx
=
∫
ln
x
d
(
ln
x
)
\int \ln xd(\ln x)
∫lnxd(lnx) 令
ln
x
\ln x
lnx为
u
u
u
=
∫
u
d
u
\int udu
∫udu
=
1
2
u
2
+
C
\frac{1}{2}u^2+C
21u2+C
=
1
2
(
ln
x
)
2
+
C
\frac{1}{2}(\ln x)^2+C
21(lnx)2+C
\quad
例题22:
∫
2
x
2
+
x
2
d
x
\int \frac{2x}{2+x^2}dx
∫2+x22xdx
=
∫
1
2
+
x
2
∗
2
x
d
x
\int \frac{1}{2+x^2}*2xdx
∫2+x21∗2xdx
=
∫
1
2
+
x
2
(
x
2
)
′
d
x
\int \frac{1}{2+x^2}(x^2)'dx
∫2+x21(x2)′dx
=
∫
1
2
+
x
2
d
(
x
2
+
2
)
\int \frac{1}{2+x^2}d(x^2+2)
∫2+x21d(x2+2)
=
ln
(
x
2
+
2
)
+
C
\ln (x^2+2)+C
ln(x2+2)+C
\quad
例题23: 若
∫
f
(
x
)
d
x
=
e
2
x
+
C
\int f(x)dx=e^{2x}+C
∫f(x)dx=e2x+C,则
f
(
x
)
=
f(x)=
f(x)=_____
2
e
2
x
2e^{2x}
2e2x
\quad
例题24:
\quad
例题25:
∫
x
1
−
x
2
d
x
\int x\sqrt{1-x^2}dx
∫x1−x2dx
=
∫
1
−
x
2
x
d
x
\int \sqrt{1-x^2}xdx
∫1−x2xdx
=
1
2
∫
1
−
x
2
2
x
d
x
\frac{1}{2}\int \sqrt{1-x^2}2xdx
21∫1−x22xdx
=
1
2
∫
1
−
x
2
(
x
2
)
′
d
x
\frac{1}{2}\int \sqrt{1-x^2}(x^2)'dx
21∫1−x2(x2)′dx
=
−
1
2
∫
1
−
x
2
d
(
−
x
2
+
1
)
-\frac{1}{2}\int \sqrt{1-x^2}d(-x^2+1)
−21∫1−x2d(−x2+1)
=
−
1
2
∗
2
3
(
1
−
x
2
)
3
2
+
C
-\frac{1}{2}*\frac{2}{3}(1-x^2)^{\frac{3}{2}}+C
−21∗32(1−x2)23+C
=
−
1
3
(
1
−
x
2
)
3
2
+
C
-\frac{1}{3}(1-x^2)^{\frac{3}{2}}+C
−31(1−x2)23+C
\quad
理解:
- [ ∫ f ( x ) d x ] ′ = f ( x ) [\int f(x)dx]'=f(x) [∫f(x)dx]′=f(x)
- d ∫ f ( x ) d x = f ( x ) d x d\int f(x)dx=f(x)dx d∫f(x)dx=f(x)dx
- ∫ F ′ ( x ) d x = F ( x ) + C \int F'(x)dx=F(x)+C ∫F′(x)dx=F(x)+C
- ∫ d F ( x ) = F ( x ) + C \int dF(x)=F(x)+C ∫dF(x)=F(x)+C \quad ∫ d sin x = sin x + C \int d\sin\sqrt{x}=\sin\sqrt{x}+C ∫dsinx=sinx+C
\quad
例题26:
∫
cos
3
x
d
x
\int \cos^3xdx
∫cos3xdx
=
∫
cos
2
x
∗
cos
x
d
x
\int \cos^2x*\cos xdx
∫cos2x∗cosxdx
=
∫
(
1
−
sin
2
x
)
(
sin
x
)
′
d
x
\int (1-\sin^2x)(\sin x)'dx
∫(1−sin2x)(sinx)′dx
=
∫
(
1
−
sin
2
x
)
d
sin
x
\int (1-\sin^2x)d\sin x
∫(1−sin2x)dsinx
=
∫
1
d
sin
x
−
∫
sin
2
x
d
sin
x
\int 1d\sin x-\int \sin^2xd\sin x
∫1dsinx−∫sin2xdsinx
=
sin
x
−
1
3
(
sin
x
)
3
+
C
\sin x-\frac{1}{3}(\sin x)^3+C
sinx−31(sinx)3+C
\quad
\quad
四. 第二类换元法
实在没办法了且带根号的才用第二换元法
第一类根换
\quad
例题27:
∫
1
1
+
x
d
x
\int \frac{1}{1+\sqrt{x}}dx
∫1+x1dx
\quad
\quad
\quad
令
x
=
u
,
x
=
u
2
\sqrt{x}=u, x=u^2
x=u,x=u2
=
∫
1
1
+
u
d
u
2
\int \frac{1}{1+u}du^2
∫1+u1du2
=
∫
1
1
+
u
(
u
2
)
′
d
u
\int \frac{1}{1+u}(u^2)'du
∫1+u1(u2)′du
=
∫
2
u
1
+
u
d
u
\int \frac{2u}{1+u}du
∫1+u2udu
=
2
∫
u
1
+
u
d
u
2\int\frac{u}{1+u}du
2∫1+uudu
=
2
∫
u
+
1
−
1
1
+
u
d
u
2\int\frac{u+1-1}{1+u}du
2∫1+uu+1−1du
=
2
∫
1
+
u
1
+
u
−
1
1
+
u
d
u
2\int\frac{1+u}{1+u}-\frac{1}{1+u}du
2∫1+u1+u−1+u1du
=
2
∫
1
d
u
−
∫
1
1
+
u
d
u
2\int1du-\int\frac{1}{1+u}du
2∫1du−∫1+u1du
=
2
u
−
ln
(
1
+
u
)
+
C
2u-\ln(1+u)+C
2u−ln(1+u)+C
=
2
x
−
ln
(
1
+
x
)
+
C
2\sqrt{x}-\ln(1+\sqrt{x})+C
2x−ln(1+x)+C
\quad
例题28:
∫
1
1
+
x
d
x
\int \frac{1}{\sqrt{1+x}}dx
∫1+x1dx
\quad
\quad
令
1
+
x
=
u
,
x
=
u
2
−
1
\sqrt{1+x}=u, \quad x=u^2-1
1+x=u,x=u2−1
=
∫
1
u
d
(
u
2
−
1
)
\int \frac{1}{u}d(u^2-1)
∫u1d(u2−1)
=
∫
1
u
d
u
2
\int \frac{1}{u}du^2
∫u1du2
=
∫
1
u
(
u
2
)
′
d
u
\int \frac{1}{u}(u^2)'du
∫u1(u2)′du
=
∫
2
u
u
d
u
\int \frac{2u}{u}du
∫u2udu
=
∫
2
d
u
\int 2du
∫2du
=
2
u
+
C
2u+C
2u+C
=
2
1
+
x
+
C
2\sqrt{1+x}+C
21+x+C
\quad
例题29:
∫
x
x
+
1
d
x
\int x\sqrt{x+1}dx
∫xx+1dx
\quad
\quad
令
x
+
1
=
u
,
x
=
u
2
−
1
\sqrt{x+1}=u, \quad x=u^2-1
x+1=u,x=u2−1
=
∫
u
(
u
2
−
1
)
d
(
u
2
−
1
)
\int u(u^2-1)d(u^2-1)
∫u(u2−1)d(u2−1)
=
∫
u
(
u
2
−
1
)
(
u
2
)
′
d
u
\int u(u^2-1)(u^2)'du
∫u(u2−1)(u2)′du
=
∫
u
(
u
2
−
1
)
2
u
d
u
\int u(u^2-1)2udu
∫u(u2−1)2udu
=
2
∫
u
4
−
u
2
d
u
2\int u^4-u^2du
2∫u4−u2du
=
2
(
1
5
u
5
−
1
3
u
3
)
+
C
2(\frac{1}{5}u^5-\frac{1}{3}u^3)+C
2(51u5−31u3)+C
=
2
5
(
x
+
1
)
5
−
2
3
(
x
+
1
)
3
+
C
\frac{2}{5}(\sqrt{x+1})^5-\frac{2}{3}(\sqrt{x+1})^3+C
52(x+1)5−32(x+1)3+C
\quad
\quad
五. 分部积分法
(
u
v
)
′
=
u
′
v
+
u
v
′
(uv)'=u'v+uv'
(uv)′=u′v+uv′
u
v
′
=
(
u
v
)
′
−
u
′
v
uv'=(uv)'-u'v
uv′=(uv)′−u′v
∫ u v ′ d x = u v − ∫ u ′ v d x \int uv'dx=uv-\int u'vdx ∫uv′dx=uv−∫u′vdx
打油诗
多指弦,
u
u
u选多
\quad
\quad
(多项式, 指数, 三角函数)
多反对,
u
u
u不选多
\quad
\quad
(多项式, 反函数, 对数)
\quad
例题30:
∫
x
ln
x
d
x
\int x\ln xdx
∫xlnxdx
\quad
\quad
(多对)
=
∫
1
2
x
2
ln
x
−
∫
1
x
∗
1
2
x
2
d
x
\int \frac{1}{2}x^2\ln x-\int \frac{1}{x}*\frac{1}{2}x^2dx
∫21x2lnx−∫x1∗21x2dx
=
∫
1
2
x
2
ln
x
−
∫
1
2
x
d
x
\int \frac{1}{2}x^2\ln x-\int \frac{1}{2}xdx
∫21x2lnx−∫21xdx
=
1
2
x
2
ln
x
−
1
4
x
2
+
C
\frac{1}{2}x^2\ln x-\frac{1}{4}x^2+C
21x2lnx−41x2+C
\quad
例题31:
∫
x
e
−
x
d
x
\int xe^{-x}dx
∫xe−xdx
\quad
\quad
(多指)
u
=
x
,
u
′
=
1
,
v
′
=
e
−
x
,
v
=
−
e
−
x
u=x, u'=1,v'=e^{-x},v=-e^{-x}
u=x,u′=1,v′=e−x,v=−e−x
=
−
x
e
−
x
−
∫
−
e
−
x
d
x
-xe^{-x}-\int -e^{-x}dx
−xe−x−∫−e−xdx
=
−
x
e
−
x
+
∫
e
−
x
d
x
-xe^{-x}+\int e^{-x}dx
−xe−x+∫e−xdx
=
−
x
e
−
x
−
e
−
x
+
C
-xe^{-x}-e^{-x}+C
−xe−x−e−x+C
\quad
例题32:
∫
ln
x
d
x
\int \ln xdx
∫lnxdx
=
x
ln
x
−
∫
1
x
∗
x
d
x
x\ln x-\int \frac{1}{x}*xdx
xlnx−∫x1∗xdx
=
x
ln
x
−
∫
1
d
x
x\ln x-\int1dx
xlnx−∫1dx
=
x
ln
x
−
x
+
C
x\ln x-x+C
xlnx−x+C
\quad
例题33:
∫
x
arctan
x
d
x
\int x \arctan xdx
∫xarctanxdx
=
1
2
x
2
arctan
x
−
1
2
∫
x
2
1
+
x
2
d
x
\frac{1}{2}x^2\arctan x-\frac{1}{2}\int \frac{x^2}{1+x^2}dx
21x2arctanx−21∫1+x2x2dx
=
1
2
x
2
arctan
x
−
1
2
∫
x
2
+
1
−
1
x
2
+
1
d
x
\frac{1}{2}x^2\arctan x-\frac{1}{2}\int \frac{x^2+1-1}{x^2+1}dx
21x2arctanx−21∫x2+1x2+1−1dx
=
1
2
x
2
arctan
x
−
1
2
(
∫
x
2
+
1
x
2
+
1
d
x
−
∫
1
x
2
+
1
d
x
)
\frac{1}{2}x^2\arctan x-\frac{1}{2}(\int \frac{x^2+1}{x^2+1}dx-\int \frac{1}{x^2+1}dx)
21x2arctanx−21(∫x2+1x2+1dx−∫x2+11dx)
=
1
2
x
2
arctan
x
−
1
2
x
+
1
2
arctan
x
+
C
\frac{1}{2}x^2\arctan x-\frac{1}{2}x+\frac{1}{2}\arctan x+C
21x2arctanx−21x+21arctanx+C
\quad
例题34:
∫
x
cos
x
d
x
\int x\cos xdx
∫xcosxdx
=
x
sin
x
−
∫
sin
x
d
x
x\sin x-\int \sin xdx
xsinx−∫sinxdx
=
x
sin
x
−
(
−
cos
x
)
+
C
x\sin x-(-\cos x)+C
xsinx−(−cosx)+C
=
x
sin
x
+
cos
x
+
C
x\sin x+\cos x+C
xsinx+cosx+C
\quad
例题35:
∫
ln
(
1
+
x
)
d
x
\int \ln(1+x)dx
∫ln(1+x)dx
=
∫
1
∗
ln
(
1
+
x
)
d
x
\int 1*\ln(1+x)dx
∫1∗ln(1+x)dx
=
x
ln
(
1
+
x
)
−
∫
x
1
+
x
d
x
x \ln(1+x)-\int \frac{x}{1+x}dx
xln(1+x)−∫1+xxdx
=
x
ln
(
1
+
x
)
−
∫
x
+
1
−
1
1
+
x
d
x
x \ln(1+x)-\int \frac{x+1-1}{1+x}dx
xln(1+x)−∫1+xx+1−1dx
=
x
ln
(
1
+
x
)
−
(
∫
1
+
x
1
+
x
d
x
−
∫
1
1
+
x
d
x
)
x \ln(1+x)-(\int \frac{1+x}{1+x}dx-\int \frac{1}{1+x}dx)
xln(1+x)−(∫1+x1+xdx−∫1+x1dx)
=
x
ln
(
1
+
x
)
−
x
+
ln
(
1
+
x
)
+
C
x \ln(1+x)-x+\ln(1+x)+C
xln(1+x)−x+ln(1+x)+C
\quad
\quad
六. 定积分
用于求曲形面积
面积只能为正, 积分有正有负
\quad
6.1. 定积分的性质
- ∫ a a f ( x ) d x = 0 \int_{a}^af(x)dx=0 ∫aaf(x)dx=0
- ∫ a b d x = ∫ a b 1 d x = b − a \int_{a}^bdx=\int_{a}^b1dx=b-a ∫abdx=∫ab1dx=b−a
- ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x \int_{a}^bf(x)dx=-\int_{b}^af(x)dx ∫abf(x)dx=−∫baf(x)dx
- ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x \int_{a}^bf(x)dx=\int_{a}^cf(x)dx+\int_{c}^bf(x)dx ∫abf(x)dx=∫acf(x)dx+∫cbf(x)dx \quad \quad 定积分的区间可加性
不论a,b,c的相对位置如何, 上式都成立
- ∫ a b [ f ( x ) ± g ( x ) ] d x = ∫ a b f ( x ) d x ± ∫ a b g ( x ) d x \int_{a}^b[f(x)\pm g(x)]dx=\int_{a}^bf(x)dx\pm \int_{a}^bg(x)dx ∫ab[f(x)±g(x)]dx=∫abf(x)dx±∫abg(x)dx \quad \quad 乘除不行
- ∫ a b k ∗ f ( x ) d x = k ∗ ∫ a b f ( x ) d x \int_{a}^bk*f(x)dx=k*\int_{a}^bf(x)dx ∫abk∗f(x)dx=k∗∫abf(x)dx
- ∫ − a a f ( x ) d x = 0 \int_{-a}^af(x)dx=0 ∫−aaf(x)dx=0 \quad \quad 奇函数在对称区间上的定积分为0
奇 ± \pm ±奇=奇 | 奇(×)(÷)奇=偶 |
---|---|
奇 ± \pm ±偶 = 非奇非偶 | 奇(×)(÷)偶=奇 |
偶 ± \pm ±(×)(÷)偶=偶 |
\quad
例题36:
(1)
∫
−
1
1
sin
x
d
x
=
0
\int_{-1}^1\sin xdx=0
∫−11sinxdx=0
(2) ∫ − 2 2 x cos x d x = 0 \int_{-2}^2x\cos xdx=0 ∫−22xcosxdx=0
(3) ∫ − 3 3 x 1 + x 2 d x = 0 \int_{-3}^3\frac{x}{\sqrt{1+x^2}}dx=0 ∫−331+x2xdx=0
\quad
例题37:
∫
−
1
1
(
2
sin
x
+
3
)
d
x
=
\int_{-1}^1(2\sin x+3)dx=
∫−11(2sinx+3)dx=_________
∫ − 1 1 2 sin x d x + ∫ − 1 1 3 d x = 0 + 3 ∗ 2 = 6 \int_{-1}^12\sin xdx+\int_{-1}^13dx=0+3*2=6 ∫−112sinxdx+∫−113dx=0+3∗2=6
\quad
例题38:
∫
−
1
1
(
5
sin
2023
x
−
2
tan
x
+
3
)
d
x
=
\int_{-1}^1(5\sin^{2023}x-2\tan x+3)dx=
∫−11(5sin2023x−2tanx+3)dx=________
∫
−
1
1
5
sin
2023
x
d
x
−
∫
−
1
1
2
tan
x
d
x
+
∫
−
1
1
3
d
x
=
0
−
0
+
6
\int_{-1}^15\sin^{2023}xdx-\int_{-1}^12\tan xdx+\int_{-1}^13dx=0-0+6
∫−115sin2023xdx−∫−112tanxdx+∫−113dx=0−0+6
\quad
\quad
七. 牛顿-莱布尼兹公式(N-L)
\quad
例题39:
∫
0
1
x
2
d
x
=
\int_0^1x^2dx=
∫01x2dx=_____
1
3
x
3
∣
0
1
=
1
3
−
0
=
1
3
\frac{1}{3}x^3|_0^1=\frac{1}{3}-0=\frac{1}{3}
31x3∣01=31−0=31
\quad
例题40:
∫
−
1
1
19
x
18
(
1
+
sin
x
)
d
x
=
\int_{-1}^119x^{18}(1+\sin x)dx=
∫−1119x18(1+sinx)dx=______
∫
−
1
1
19
x
18
d
x
+
∫
−
1
1
sin
x
d
x
=
19
∗
1
19
x
19
∣
−
1
1
+
0
=
1
−
(
−
1
)
=
2
\int_{-1}^119x^{18}dx+\int_{-1}^1\sin xdx=19*\frac{1}{19}x^{19}|_{-1}^1+0=1-(-1)=2
∫−1119x18dx+∫−11sinxdx=19∗191x19∣−11+0=1−(−1)=2
\quad
例题41:
f
(
x
)
=
{
x
x
≤
1
2
x
x
>
1
f(x)=\begin{cases} x & x\leq1 \\ 2x & x>1 \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \end{cases}
f(x)={x2xx≤1x>1
求
∫
0
2
f
(
x
)
d
x
\int_{0}^2f(x)dx
∫02f(x)dx
=
∫
0
1
f
(
x
)
d
x
+
∫
1
2
f
(
x
)
d
x
\int_{0}^1f(x)dx+\int_{1}^2f(x)dx
∫01f(x)dx+∫12f(x)dx
=
1
2
x
2
∣
0
1
+
x
2
∣
1
2
\frac{1}{2}x^2|_0^1+x^2|_1^2
21x2∣01+x2∣12
=
3
1
2
3\frac{1}{2}
321
\quad
例题42:
∫
0
1
(
2
x
−
1
)
5
d
x
=
\int_{0}^1(2x-1)^5dx=
∫01(2x−1)5dx=_______
=
1
2
∫
0
1
(
2
x
−
1
)
5
d
(
2
x
)
\frac{1}{2}\int_{0}^1(2x-1)^5d(2x)
21∫01(2x−1)5d(2x)
=
1
2
∫
0
1
(
2
x
−
1
)
5
d
(
2
x
−
1
)
\frac{1}{2}\int_{0}^1(2x-1)^5d(2x-1)
21∫01(2x−1)5d(2x−1)
=
1
2
∗
1
6
(
2
x
−
1
)
6
∣
0
1
\frac{1}{2}*\frac{1}{6}(2x-1)^6|_0^1
21∗61(2x−1)6∣01
=
0
0
0
\quad
\quad
\quad
例题43:
∫
−
2
−
1
1
1
+
2
x
+
5
d
x
\int_{-2}^{-1}\frac{1}{1+\sqrt{2x+5}}dx
∫−2−11+2x+51dx
\quad
\quad
八. 定积分的分部积分法
\quad
\quad
例题44:
∫
0
π
2
x
sin
x
d
x
\int_0^{\frac{π}{2}}x\sin xdx
∫02πxsinxdx
=
[
0
−
0
]
−
[
−
1
−
0
]
=
1
[0-0]-[-1-0]=1
[0−0]−[−1−0]=1
\quad
例题45:
=
1
1
1
\quad
例题46:
=
e
2
4
+
1
4
\frac{e^2}{4}+\frac{1}{4}
4e2+41
\quad
例题47:
\quad
例题48:
\quad
例题49:
∫
−
1
2
x
e
x
2
d
x
\int_{-1}^2xe^{x^2}dx
∫−12xex2dx
=
1
2
∫
−
1
2
e
x
2
∗
2
x
d
x
\frac{1}{2}\int_{-1}^2e^{x^2}*2xdx
21∫−12ex2∗2xdx
=
1
2
∫
−
1
2
e
x
2
∗
(
x
2
)
′
d
x
\frac{1}{2}\int_{-1}^2e^{x^2}*(x^2)'dx
21∫−12ex2∗(x2)′dx
=
1
2
∫
−
1
2
e
x
2
d
(
x
2
)
\frac{1}{2}\int_{-1}^2e^{x^2}d(x^2)
21∫−12ex2d(x2)
=
e
4
2
−
e
2
\frac{e^4}{2}-\frac{e}{2}
2e4−2e
\quad
\quad
九. 定积分的几何应用
9.1 平面图形的面积
\quad
例50: 计算由直线
y
=
x
y=x
y=x和抛物线
y
=
x
2
y=x^2
y=x2所围成的平面图形的面积
交点(0,1), (1,1)
S 夹 = ∫ 0 1 x d x − ∫ 0 1 x 2 d x = 1 2 x 2 ∣ 0 1 − 1 3 x 3 ∣ 0 1 = 1 6 S_夹=\int_0^1xdx-\int_0^1x^2dx=\frac{1}{2}x^2|_0^1-\frac{1}{3}x^3|_0^1=\frac{1}{6} S夹=∫01xdx−∫01x2dx=21x2∣01−31x3∣01=61
\quad
\quad
例51: 计算由直线
y
=
x
+
2
y=x+2
y=x+2和抛物线
y
=
x
2
y=x^2
y=x2所围成的平面图形的面积
{
y
=
x
+
2
y
=
x
2
\begin{cases} y=x+2 \\ y=x^2 \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\qquad\quad\qquad\qquad\qquad\qquad \end{cases}
{y=x+2y=x2
解出交点为(-1,1), (2,4)
S = ∫ − 1 2 ( x + 2 ) − x 2 d x = ( 1 2 x 2 + 2 x − 1 3 x 3 ) ∣ − 1 2 = 9 2 S=\int_{-1}^2(x+2)-x^2dx=(\frac{1}{2}x^2+2x-\frac{1}{3}x^3)|_{-1}^2=\frac{9}{2} S=∫−12(x+2)−x2dx=(21x2+2x−31x3)∣−12=29
\quad
\quad
9.2 旋转体的体积
\quad
\quad
例52: 设曲线
y
=
e
x
y=e^x
y=ex, 与直线
x
=
0
,
x
=
1
x=0, x=1
x=0,x=1和
x
x
x轴围成的平面图形为D
(1) 求D的面积
(2) 求D围绕x轴旋转一周所得旋转体的体积
\quad
例53: 一直平面图形D由曲线
y
=
e
x
,
y
=
x
,
x
=
0
,
x
=
1
y=e^x, y=x, x=0, x=1
y=ex,y=x,x=0,x=1围成
(1) 求D的面积A
(2) 求D绕x轴旋转一周所得的体积V
\quad
例54: 设曲线
y
=
cos
x
(
x
∈
[
0
,
π
2
]
)
y=\cos x(x\in [0,\frac{π}{2}])
y=cosx(x∈[0,2π])与x轴及y轴所围成的平面图形为D
(1) 求D的面积A
(1) 求D绕x轴旋转一周所得旋转体的体积
\quad
例55: 已知平面图形D由曲线
y
=
1
+
x
,
y
=
1
+
x
y=\sqrt{1+x}, y=1+x
y=1+x,y=1+x围成
(1) 求D的面积A
(1) 求D绕x轴旋转一周所得旋转体的体积
\quad
\quad
十. 广义积分
\quad
例:
∫
0
1
1
x
d
x
=
\int_0^1\frac{1}{\sqrt{x}}dx=
∫01x1dx=_______
∫ 0 1 x − 1 2 d x = 2 x 1 2 ∣ 0 1 = 2 \int_0^1 x^{-\frac{1}{2}}dx=2x^{\frac{1}{2}}|_0^1=2 ∫01x−21dx=2x21∣01=2
\quad
例:
∫
−
∞
1
e
x
d
x
=
\int_{-\infty}^1e^xdx=
∫−∞1exdx=______
e x ∣ − ∞ 1 = e − 0 = e e^x|_{-\infty}^1=e-0=e ex∣−∞1=e−0=e
\quad
例:
∫
−
∞
+
∞
1
1
+
x
2
d
x
\int_{-\infty}^{+\infty}\frac{1}{1+x^2}dx
∫−∞+∞1+x21dx______
arctan x ∣ − ∞ + ∞ = π 2 − ( − π 2 ) = π \arctan x|_{-\infty}^{+\infty}=\frac{π}{2}-(-\frac{π}{2})=π arctanx∣−∞+∞=2π−(−2π)=π
\quad
\quad
十一. 变上限积分
\quad
例: 设函数
f
(
x
)
=
∫
0
x
cos
2
t
d
t
f(x)=\int_0^x\cos^2tdt
f(x)=∫0xcos2tdt, 则
f
′
(
x
)
=
f'(x)=
f′(x)=
cos
2
x
\cos^2x
cos2x文章来源:https://www.toymoban.com/news/detail-786001.html
\quad
例: 设函数
f
(
x
)
=
∫
0
x
1
1
+
t
2
d
t
f(x)=\int_0^x\frac{1}{1+t^2}dt
f(x)=∫0x1+t21dt, 则
f
′
(
x
)
=
f'(x)=
f′(x)=
1
1
+
x
2
\frac{1}{1+x^2}
1+x21文章来源地址https://www.toymoban.com/news/detail-786001.html
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