常见函数积分 |
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∫
k
d
x
=
k
x
+
C
\int k\text{d}x=kx+C
∫kdx=kx+C
∫
x
μ
d
x
=
x
μ
+
1
μ
+
1
+
C
(
μ
≠
1
)
\int x^{\mu}\text{d}x=\frac{x^{\mu+1}}{\mu+1}+C(\mu\neq 1)
∫xμdx=μ+1xμ+1+C(μ=1)
∫
d
x
x
=
ln
∣
x
∣
+
C
\int \frac{\text{d}x}{x}=\text{ln}|x|+C
∫xdx=ln∣x∣+C
∫
d
x
1
+
x
2
=
a
r
c
t
a
n
x
+
C
\int \frac{\text{d}x}{1+x^2}=arctan\ x+C
∫1+x2dx=arctan x+C
∫
d
x
1
−
x
2
=
a
r
c
s
i
n
x
+
C
\int \frac{\text{d}x}{\sqrt {1-x^2}}=arcsin\ x+C
∫1−x2
dx=arcsin x+C
∫
c
o
s
x
d
x
=
s
i
n
x
+
C
\int cos\ x\text{d}x=sin\ x+C
∫cos xdx=sin x+C
∫
s
i
n
x
d
x
=
−
c
o
s
x
+
C
\int sin\ x\text{d}x=-cos\ x+C
∫sin xdx=−cos x+C
∫
d
x
c
o
s
2
x
=
t
a
n
x
+
C
\int \frac{\text{d}x}{cos^2\ x}=tan\ x+C
∫cos2 xdx=tan x+C
∫
d
x
s
i
n
2
x
=
−
c
o
t
x
+
C
\int \frac{\text{d}x}{sin^2\ x}=-cot\ x+C
∫sin2 xdx=−cot x+C
∫
s
e
c
x
t
a
n
x
d
x
=
s
e
c
x
+
C
\int sec\ xtan\ x\text{d}x=sec\ x+C
∫sec xtan xdx=sec x+C
∫
c
s
c
x
c
o
t
x
d
x
=
−
c
s
c
x
+
C
\int csc\ xcot\ x\text{d}x=-csc\ x+C
∫csc xcot xdx=−csc x+C
∫
e
x
d
x
=
e
x
+
C
\int e^x\text{d}x=e^x+C
∫exdx=ex+C
∫
a
x
d
x
=
a
x
ln
a
+
C
\int a^x\text{d}x=\frac{a^x}{\text{ln}\ a}+C
∫axdx=ln aax+C
∫
t
a
n
x
d
x
=
−
ln
∣
c
o
s
x
∣
+
C
\int tan\ x\text{d}x=-\text{ln}\ |cos\ x|+C
∫tan xdx=−ln ∣cos x∣+C
∫
c
o
t
x
d
x
=
ln
∣
s
i
n
x
∣
+
C
\int cot\ x\text{d}x=\text{ln}\ |sin\ x|+C
∫cot xdx=ln ∣sin x∣+C
∫
d
x
a
2
+
x
2
=
1
a
a
r
c
t
a
n
x
a
+
C
\int \frac{\text{d}x}{a^2+x^2}=\frac{1}{a}arctan\ \frac{x}{a}+C
∫a2+x2dx=a1arctan ax+C
∫
d
x
a
2
−
x
2
=
a
r
c
s
i
n
x
a
+
C
\int \frac{\text{d}x}{\sqrt{a^2-x^2}}=arcsin\ \frac{x}{a}+C
∫a2−x2
dx=arcsin ax+C
∫
d
x
a
2
+
x
2
=
ln
(
x
+
a
2
+
x
2
)
+
C
\int \frac{\text{d}x}{\sqrt{a^2+x^2}}=\text{ln}\ (x+\sqrt{a^2+x^2})+C
∫a2+x2
dx=ln (x+a2+x2
)+C
∫
d
x
x
2
−
a
2
=
ln
∣
x
+
x
2
−
a
2
∣
+
C
\int\frac{\text{d}x}{\sqrt{x^2-a^2}}=\text{ln}\ |x+\sqrt{x^2-a^2}|+C
∫x2−a2
dx=ln ∣x+x2−a2
∣+C
第一类换元法： 找到 f ( x ) = g [ φ ( x ) ] φ ′ ( x ) f(x)=g[\varphi(x)]\varphi'(x) f(x)=g[φ(x)]φ′(x) ∫ f ( x ) d x = ∫ g [ φ ( x ) ] φ ′ ( x ) d x = ∫ g [ φ ( x ) ] d φ ( x ) = [ ∫ g ( u ) d u ] u = φ ( x ) \int f(x)\text{d}x=\int g[\varphi(x)]\varphi'(x)\text{d}x=\int g[\varphi(x)]\text{d}\varphi(x)=[\int g(u)\text{d}u]_{u=\varphi(x)} ∫f(x)dx=∫g[φ(x)]φ′(x)dx=∫g[φ(x)]dφ(x)=[∫g(u)du]u=φ(x) 第二类换元法： 找到 x = φ ( t ) x=\varphi(t) x=φ(t) ∫ f ( x ) d x = [ ∫ f ( φ ( t ) ) φ ′ ( t ) d t ] t = φ − 1 ( x ) \int f(x)\text{d}x=[\int f(\varphi(t))\varphi'(t)\text{d}t]_{t=\varphi^{-1}(x)} ∫f(x)dx=[∫f(φ(t))φ′(t)dt]t=φ−1(x) |
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