Proprietăţi ale integralelor nedefinite
modificare
a) Dacă
f
:
I
→
R
{\displaystyle f:I\to \mathbb {R} }
admite primitive pe I , iar
a
∈
R
,
{\displaystyle a\in \mathbb {R} ,}
atunci şi
a
f
{\displaystyle \mathrm {a} f}
admite primitive şi:
∫
a
f
(
x
)
d
x
=
a
∫
f
(
x
)
d
x
.
{\displaystyle \int af(x)\,\mathrm {d} x=a\int f(x)\,\mathrm {d} x.}
b) Dacă f şi g admit primitive pe I atunci
f
+
g
{\displaystyle f+g}
admite primitive pe I şi:
∫
(
f
(
x
)
+
g
(
x
)
)
d
x
=
∫
f
(
x
)
d
x
+
∫
g
(
x
)
d
x
{\displaystyle \int \left(f(x)+g(x)\right)\,\mathrm {d} x=\int f(x)\mathrm {d} x+\int g(x)\mathrm {d} x}
Integrale nedefinite utilizate frecvent
modificare
f
:
R
→
R
,
f
(
x
)
=
x
n
,
n
∈
N
{\displaystyle f:\mathbb {R} \to \mathbb {R} ,\;f(x)=x^{n},\;n\in \mathbb {N} \;\;}
∫
x
n
d
x
=
x
n
+
1
n
+
1
+
C
.
{\displaystyle \int x^{n}\mathrm {d} x={\frac {x^{n+1}}{n+1}}+{\mathcal {C}}.}
f
:
I
→
R
,
I
⊂
(
0
,
∞
)
,
f
(
x
)
=
x
a
,
a
∈
R
∖
{
1
}
{\displaystyle f:I\to \mathbb {R} ,\;I\subset (0,\infty )\;,\;f(x)=x^{a}\;,\;a\in \mathbb {R} \setminus \{1\}}
∫
x
a
d
x
=
x
a
+
1
a
+
1
+
C
.
{\displaystyle \int x^{a}\mathrm {d} x={\frac {x^{a+1}}{a+1}}+{\mathcal {C}}.}
f
:
I
→
R
,
I
⊂
R
∗
,
f
(
x
)
=
1
x
{\displaystyle f:I\to \mathbb {R} \;,\;I\subset \mathbb {R} ^{*}\;,\;f(x)={\frac {1}{x}}}
∫
1
x
d
x
=
ln
|
x
|
+
C
.
{\displaystyle \int {\frac {1}{x}}\mathrm {d} x=\ln |x|+{\mathcal {C}}.}
f
:
R
→
R
,
f
(
x
)
a
x
,
a
>
0
,
a
≠
1
{\displaystyle f:\mathbb {R} \to \mathbb {R} \;,\;f(x)a^{x}\;,\;a>0\;,\;a\neq 1}
∫
a
x
d
x
=
a
x
ln
a
+
C
.
{\displaystyle \int a^{x}\mathrm {d} x={\frac {a^{x}}{\ln a}}+{\mathcal {C}}.}
f
:
I
→
R
,
I
⊂
R
∖
{
−
a
,
a
}
,
f
(
x
)
=
1
x
2
−
a
2
,
a
≠
0
{\displaystyle f:I\to \mathbb {R} \;,\;I\subset \mathbb {R} \setminus \{-a,a\}\,,\,f(x)={\frac {1}{x^{2}-a^{2}}}\,,\,a\neq 0}
∫
d
x
x
2
−
a
2
=
1
2
a
ln
|
x
−
a
x
+
a
|
+
C
.
{\displaystyle \int {\frac {\mathrm {d} x}{x^{2}-a^{2}}}={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+{\mathcal {C}}.}
{\displaystyle }
{\displaystyle }
{\displaystyle }
{\displaystyle }