高等数学积分表公式推导

∴
∫
dx ax +
b
=
1 a
∫
1dt t
= 1 ⋅ ln t + C a
将
t
=
ax
+
b
代入上式得：∫
dx ax +
b
=
1 a
⋅ ln
ax + b
+C
∫ 2. (ax + b)μ dx = 1 ⋅ (ax + b)μ+1 + C (μ ≠ −1)
a (μ + 1)
证明：令 ax + b = t ,则dt = adx ,∴ dx = 1 dt a
=
1 a2
(t −
b⋅ ln
t
)+ C
将t
=
ax +
b 代入上式得：∫
x ax +
b
dx
=
1 a2
(ax+
b−
b⋅ ln
ax+ b
)+ C
-1-
∫ 4.
x2 dx
ax + b
=
1 a3
⎡ ⎢⎣
1 2
(ax
+
b)
2
−
2b (ax + b)
+ b2
⋅ ln
ax + b
⎤ ⎥⎦
+
C
∫ ∫ x2
1
证明：
ax
高等数学
积分表 公式推导
目录
（一）含有 ax + b 的积分（1~9）·······················································1 （二）含有 ax + b 的积分（10~18）···················································5 （三）含有 x 2 ± a 2 的积分（19~21）····················································9 （四）含有 ax2 + b (a > 0) 的积分（22~28）············································11 （五）含有 ax2 + bx + c (a > 0) 的积分（29~30）········································14 （六）含有 x 2 + a 2 (a > 0) 的积分（31~44）·········································15 （七）含有 x 2 − a 2 (a > 0) 的积分（45~58）·········································24 （八）含有 a 2 − x 2 (a > 0) 的积分（59~72）·········································37 （九）含有 ± a 2 + bx + c (a > 0) 的积分（73~78）····································48 （十）含有 ± x − a 或 ( x − a )( b − x ) 的积分（79~82）···························51
+
dx b
=
a2
(ax + b)2 − 2abx − b2 )dx ax + b
=
1 a2
∫
(ax
+ b)dx
−
1 a2
∫
2abx dx ax + b
−
1 a2
∫
b2 ax +
dx b
∫ ∵ 1
a2
(ax +
b)dx
=
1 2a 3
(ax
+
b) 2
+ C1
1 a2
∫
2abx dx ax + b
=
2b a3
附录：常数和基本初等函数导数公式·········································85
（一）含有 ax + b 的积分（1~9）
1.
∫
dx ax +
b
=
1 a
⋅ ln
ax + b
+C
证明：被积函数 f ( x ) =
1
的定义域为{ x
|
x
≠
−
b }
ax + b
a
令 ax + b = t (t ≠ 0)，则dt = adx ,∴ dx = 1 dt a
x−b
（十一）含有三角函数的积分（83~112）···········································55 （十二）含有反三角函数的积分（其中 a > 0)（113~121）·······················68 （十三）含有指数函数的积分（122~131）··········································73 （十四）含有对数函数的积分（132~136）··········································78 （十五）含有双曲函数的积分（137~141）··········································80 （十六）定积分（142~147）····························································81
ax +
dx b
=
1 a3
⎡1 ⎢⎣ 2
(ax +
b) 2
−
2b (ax
+
b)
+
b2
⋅ ln
ax +
b
⎤ ⎥⎦
+C
5.
dx
1
∫ x (ax + b) = − b ⋅ ln
ax + b x
+C
证明：被积函数 f ( x ) = 1 的定义域为{x | x ≠ − b}
x ⋅ (ax+ b)
∫
ax + b ax +
− bd(ax) b
2b
2b2 1
= a3 ∫ dx −
a3
∫
ax
+
d b
(ax
+
b)
=
2b a3
x
−
2b2 a3
ln
ax + b
+ C2
∫ ∫ 1
a2
b2 ax +
dx b
=
b2 a3
1 ax +
d (ax b
+
b)
=
b2 a3
ln
ax + b
+ C3
∫ 由以上各式整理得： x2
∴
∫
( ax
+
b) μdx
=
1 a
∫t
μ dt
= 1 ⋅ t μ+1 + C a (μ + 1)
∫ 将t = ax + b代入上式得：(ax + b) μ dx = 1 ⋅ (ax + b) μ+1 + C
a (μ + 1)
3.
∫
x ax +
b
dx
=
1 a2
(ax+
b−
b⋅ ln
ax+ b
)+ C
证明：被积函数 f ( x ) = x 的定义域为 {x | x ≠ − b }
ax+ b
a
令 ax+ b = t (t ≠ 0), 则 x = 1 (t − b) , dx = 1 dt
a
a
∴
∫
x ax+
bห้องสมุดไป่ตู้
dx
=∫
1 a
(t −
t
b )· 1
a
dt
=
1 a2
∫
⎜⎛1 ⎝
−
b t
⎟⎞ dt ⎠
=
1 a2
∫ dt−
1 a2
b
∫t
dt
tb = a 2 − a 2 ⋅ ln t + C