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Sunday, 14 October 2018
Danh sách tích phân với hàm mũ – Wikipedia tiếng Việt
Dưới đây là
danh sách các tích phân với hàm mũ
.
∫
e
c
x
d
x
=
1
c
e
c
x
{displaystyle int e^{cx};dx={frac {1}{c}}e^{cx}}
∫
a
c
x
d
x
=
1
c
ln
a
a
c
x
(
a
>
0
,
a
≠
1
)
{displaystyle int a^{cx};dx={frac {1}{cln a}}a^{cx}qquad {mbox{(}}a>0,{mbox{ }}aneq 1{mbox{)}}}
∫
x
e
c
x
d
x
=
e
c
x
c
2
(
c
x
−
1
)
{displaystyle int xe^{cx};dx={frac {e^{cx}}{c^{2}}}(cx-1)}
∫
x
2
e
c
x
d
x
=
e
c
x
(
x
2
c
−
2
x
c
2
+
2
c
3
)
{displaystyle int x^{2}e^{cx};dx=e^{cx}left({frac {x^{2}}{c}}-{frac {2x}{c^{2}}}+{frac {2}{c^{3}}}right)}
∫
x
n
e
c
x
d
x
=
1
c
x
n
e
c
x
−
n
c
∫
x
n
−
1
e
c
x
d
x
{displaystyle int x^{n}e^{cx};dx={frac {1}{c}}x^{n}e^{cx}-{frac {n}{c}}int x^{n-1}e^{cx}dx}
∫
e
c
x
d
x
x
=
ln
|
x
|
+
∑
i
=
1
∞
(
c
x
)
i
i
⋅
i
!
{displaystyle int {frac {e^{cx};dx}{x}}=ln |x|+sum _{i=1}^{infty }{frac {(cx)^{i}}{icdot i!}}}
∫
e
c
x
d
x
x
n
=
1
n
−
1
(
−
e
c
x
x
n
−
1
+
c
∫
e
c
x
x
n
−
1
d
x
)
(
n
≠
1
)
{displaystyle int {frac {e^{cx};dx}{x^{n}}}={frac {1}{n-1}}left(-{frac {e^{cx}}{x^{n-1}}}+cint {frac {e^{cx}}{x^{n-1}}},dxright)qquad {mbox{(}}nneq 1{mbox{)}}}
∫
e
c
x
ln
x
d
x
=
1
c
e
c
x
ln
|
x
|
−
Ei
(
c
x
)
{displaystyle int e^{cx}ln x;dx={frac {1}{c}}e^{cx}ln |x|-operatorname {Ei} ,(cx)}
∫
e
c
x
sin
b
x
d
x
=
e
c
x
c
2
+
b
2
(
c
sin
b
x
−
b
cos
b
x
)
{displaystyle int e^{cx}sin bx;dx={frac {e^{cx}}{c^{2}+b^{2}}}(csin bx-bcos bx)}
∫
e
c
x
cos
b
x
d
x
=
e
c
x
c
2
+
b
2
(
c
cos
b
x
+
b
sin
b
x
)
{displaystyle int e^{cx}cos bx;dx={frac {e^{cx}}{c^{2}+b^{2}}}(ccos bx+bsin bx)}
∫
e
c
x
sin
n
x
d
x
=
e
c
x
sin
n
−
1
x
c
2
+
n
2
(
c
sin
x
−
n
cos
x
)
+
n
(
n
−
1
)
c
2
+
n
2
∫
e
c
x
sin
n
−
2
x
d
x
{displaystyle int e^{cx}sin ^{n}x;dx={frac {e^{cx}sin ^{n-1}x}{c^{2}+n^{2}}}(csin x-ncos x)+{frac {n(n-1)}{c^{2}+n^{2}}}int e^{cx}sin ^{n-2}x;dx}
∫
e
c
x
cos
n
x
d
x
=
e
c
x
cos
n
−
1
x
c
2
+
n
2
(
c
cos
x
+
n
sin
x
)
+
n
(
n
−
1
)
c
2
+
n
2
∫
e
c
x
cos
n
−
2
x
d
x
{displaystyle int e^{cx}cos ^{n}x;dx={frac {e^{cx}cos ^{n-1}x}{c^{2}+n^{2}}}(ccos x+nsin x)+{frac {n(n-1)}{c^{2}+n^{2}}}int e^{cx}cos ^{n-2}x;dx}
∫
x
e
c
x
2
d
x
=
1
2
c
e
c
x
2
{displaystyle int xe^{cx^{2}};dx={frac {1}{2c}};e^{cx^{2}}}
∫
1
σ
2
π
e
−
(
x
−
μ
)
2
/
2
σ
2
d
x
=
1
2
σ
(
1
+
erf
s="MJX-TeXAtom-ORD">
x
−
μ
σ
2
)
{displaystyle int {1 over sigma {sqrt {2pi }}},e^{-{(x-mu )^{2}/2sigma ^{2}}};dx={frac {1}{2sigma }}(1+{mbox{erf}},{frac {x-mu }{sigma {sqrt {2}}}})}
∫
e
x
2
d
x
=
e
x
2
(
∑
j
=
0
n
−
1
c
2
j
1
x
2
j
+
1
)
+
(
2
n
−
1
)
c
2
n
−
2
∫
e
x
2
x
2
n
d
x
(
n
>
0
)
,
{displaystyle int e^{x^{2}},dx=e^{x^{2}}left(sum _{j=0}^{n-1}c_{2j},{frac {1}{x^{2j+1}}}right)+(2n-1)c_{2n-2}int {frac {e^{x^{2}}}{x^{2n}}};dxquad (n>0),}
với
Không thể phân tích cú pháp (lỗi cú pháp): {displaystyle c_{2j}=frac{ 1 cdot 3 cdot 5 cdots (2j-1)}{2^{j+1}}=frac{(2j),!}{j!, 2^{2j+1}} . }
∫
−
∞
∞
e
−
a
x
2
d
x
=
π
a
{displaystyle int _{-infty }^{infty }e^{-ax^{2}},dx={sqrt {pi over a}}}
∫
0
∞
x
2
n
e
−
x
2
/
a
2
d
x
=
π
(
2
n
)
!
n
!
(
a
2
)
2
n
+
1
{displaystyle int _{0}^{infty }x^{2n}e^{-{x^{2}}/{a^{2}}},dx={sqrt {pi }}{(2n)! over {n!}}{left({frac {a}{2}}right)}^{2n+1}}
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