BẢNG TÍCH PHÂN
∫ xα = 1
1+α ∗ x1+α + C
∫ dx x =ln │ x │+ C
∫ axdx = ax/lna + C
∫ eax = 1
aeax + C
∫ cosbxdx = 1
∫ sinbxdx = ư1
∫ cos2 x dx = tgx+C
∫ sin 2 x dx = cotgx+C
∫ sinx dx =ln │tg ( 2 x ) │+C
∫ cosx dx = ln │ tg ( x 2 +
Π
4 ) │+C
∫ a 2ưx 2 dx = 1
2 a ln │
a+ x aưx │+C
∫ a 2+x 2 dx = 1
a arctag │
x
a │+C
∫ dx
√ x 2+b = ln│x + √ x 2+b│ + C
∫ √ aưx 2 dx = x
2 √ a 2ưx 2 +
a 2
2 arcsin
x
a + C
∫ √ x 2+ b dx = x
2 √ x 2+b +
b
2 ln│x + √ x 2+b│ + C
∫ udv=uvư ∫ vdu
∫
0
Π / 2
sin nxdx = ∫
0
Π / 2
cos nxdx
= (nư1) ‼
2
(n+1)‼
CÔNG THỨC TÍNH VI PHÂN
(uα)’ = α uα-1 .u’ => u’ => ¿)’ =(1/2 √ u).u’ => u’
(au)’ = au.u’ => lna u’ => u’ (eu)’ = eu.u’ => u’
(ln u)’ = u'
1
ulna.u’ => u’
(sin u)’ = cos u u’ => u’ (cos u)’ = -sin u u’ => u’ (tag u)’ = 1/cos2u u’ => u’ (cotag u)’ = 1/sin2u u’ => u’ (arcsin u)’ = 1/√ 1ưu 2 u’ => u’
(arccos u)’ = -1/√ 1ưu 2 u’ => u’
(arctag u)’ = 1/(1+u2) u’ => u’
(arccotag u)’ = -1/(1+u2) u’ => u’
(y)’ = (u(x)v(x))’ = uv [v’ln u + v
u u’ => u’]
Trang 2BẢNG TÍCH PHÂN
∫ xα = 1
1+α ∗ x1+α + C
∫ dx x =ln │ x │+ C
∫ axdx = ax/lna + C
∫ eax = 1
aeax + C
∫ cosbxdx = 1
∫ sinbxdx = ư1
∫ cos2 x dx = tgx+C
∫ sin 2 x dx = cotgx+C
∫ sinx dx =ln │tg ( 2 x ) │+C
∫ cosx dx = ln │ tg ( x 2 +
Π
4 ) │+C
∫ a 2ưx 2 dx = 1
2 a ln │
a+ x aưx │+C
∫ a 2+x 2 dx = 1
a arctag │
x
a │+C
∫ dx
√ x 2+b = ln│x + √ x 2+b│ + C
∫ √ aưx 2 dx = x
2 √ a 2ưx 2 +
a 2
2 arcsin
x
a + C
∫ √ x 2+b dx = x
2 √ x 2+b +
b
2 ln│x + √ x 2+b│ + C
∫ udv=uvư ∫ vdu
∫
0
Π / 2
sin nxdx = ∫
0
Π / 2
cos nxdx
= (nư1) ‼
2
(n+1)‼
CÔNG THỨC TÍNH VI PHÂN
(uα)’ = α uα-1 .u’ => u’ => ¿)’ =(1/2 √ u).u’ => u’
(au)’ = au.u’ => lna u’ => u’ (eu)’ = eu.u’ => u’
(ln u)’ = u'
1
ulna.u’ => u’
(sin u)’ = cos u u’ => u’ (cos u)’ = -sin u u’ => u’
(tag u)’ = 1/cos2u u’ => u’ (cotag u)’ = 1/sin2u u’ => u’ (arcsin u)’ = 1/√ 1ưu 2 u’ => u’
(arccos u)’ = -1/√ 1ưu 2 u’ => u’
(arctag u)’ = 1/(1+u2) u’ => u’
(arccotag u)’ = -1/(1+u2) u’ => u’
Trang 3(y)’ = (u(x)v(x))’ = uv [v’ln u + v
u u’ => u’]
BẢNG TÍCH PHÂN
∫ xα = 1
1+α ∗ x1+α + C
∫ dx x =ln │ x │+ C
∫ axdx = ax/lna + C
∫ eax = 1
aeax + C
∫ cosbxdx = 1
∫ sinbxdx = −1
∫ cos2 x dx = tgx+C
∫ sin 2 x dx = cotgx+C
∫ sinx dx =ln │tg ( 2 x ) │+C
∫ cosx dx =ln │ tg ( x 2 +
Π
4 ) │+C
∫ a 2−x 2 dx = 1
2 a ln │
a+ x a−x │+C
∫ a 2+x 2 dx = 1
a arctag │
x
a │+C
∫ dx
√ x 2+b = ln│x + √ x 2+b│ + C
∫ √ a−x 2 dx = x
2 √ a 2−x 2 +
a 2
2 arcsin
x
a + C
∫ √ x 2+ b dx = x
2 √ x 2+b +
b
2 ln│x + √ x 2+b│ + C
∫ udv=uv− ∫ vdu
∫
0
Π / 2
sin nxdx = ∫
0
Π / 2
cos nxdx
= (n−1) ‼
2
(n+1)‼
CÔNG THỨC TÍNH VI PHÂN
(uα)’ = α uα-1 .u’ => u’ => ¿)’ =(1/2 √ u).u’ => u’
(au)’ = au.u’ => lna u’ => u’ (eu)’ = eu.u’ => u’
(ln u)’ = u'
1
ulna.u’ => u’
Trang 4(sin u)’ = cos u u’ => u’ (cos u)’ = -sin u u’ => u’
(tag u)’ = 1/cos2u u’ => u’ (cotag u)’ = 1/sin2u u’ => u’
(arcsin u)’ = 1/√ 1ưu 2 u’ => u’
(arccos u)’ = -1/√ 1ưu 2 u’ => u’
(arctag u)’ = 1/(1+u2) u’ => u’
(arccotag u)’ = -1/(1+u2) u’ => u’
(y)’ = (u(x)v(x))’ = uv [v’ln u + v
u u’ => u’]
BẢNG TÍCH PHÂN
∫ xα = 1
1+α ∗ x1+α + C
∫ dx x =ln │ x │+ C
∫ axdx = ax/lna + C
∫ eax = 1
aeax + C
∫ cosbxdx = 1
∫ sinbxdx = ư1
∫ cos2 x dx = tgx+C
∫ sin 2 x dx = cotgx+C
∫ sinx dx =ln │tg ( 2 x ) │+C
∫ cosx dx =ln │ tg ( x 2 +
Π
4 ) │+C
∫ a 2ưx 2 dx = 1
2 a ln │
a+ x aưx │+C
∫ a 2+x 2 dx = 1
a arctag │
x
a │+C
∫ dx
√ x 2+b = ln│x + √ x 2+b│ + C
∫ √ aưx 2 dx = x
2 √ a 2ưx 2 +
a 2
2 arcsin
x
a + C
∫ √ x 2+ b dx = x
2 √ x 2+b +
b
2 ln│x + √ x 2+b│ + C
∫ udv=uvư ∫ vdu
∫
0
Π / 2
sin nxdx = ∫
0
Π / 2
cos nxdx
= (nư1) ‼
2
(n+1)‼
Trang 5CÔNG THỨC TÍNH VI PHÂN
(uα)’ = α uα-1 .u’ => u’ => ¿)’ =(1/2 √ u).u’ => u’
(au)’ = au.u’ => lna u’ => u’ (eu)’ = eu.u’ => u’
(ln u)’ = u'
1
ulna.u’ => u’
(sin u)’ = cos u u’ => u’ (cos u)’ = -sin u u’ => u’ (tag u)’ = 1/cos2u u’ => u’ (cotag u)’ = 1/sin2u u’ => u’ (arcsin u)’ = 1/√ 1ưu 2 u’ => u’
(arccos u)’ = -1/√ 1ưu 2 u’ => u’
(arctag u)’ = 1/(1+u2) u’ => u’
(arccotag u)’ = -1/(1+u2) u’ => u’
(y)’ = (u(x)v(x))’ = uv [v’ln u + v
u u’ => u’]