# Primitive

Exerciții și probleme... primitive.

Funcția
Mulțimea primitivelor
$f : \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = x^n$,
unde $n \in \mathbb{N}$
$\displaystyle \int x^n dx =$ $\displaystyle \frac{x^{n+1}}{n+1} +C$
$f : I \rightarrow \mathbb{R}$, $f(x) = x^{\alpha}$,
unde $I \subset (0, \infty)$, $\alpha \in \mathbb{R} \setminus \{ -1 \}$
$\displaystyle \int x^{\alpha} dx =$ $\displaystyle \frac{x^{\alpha+1}}{\alpha+1} +C$
$f : I \rightarrow \mathbb{R}$, $\displaystyle f(x) = \frac{1}{x}$,
unde $I \subset (0, \infty)$ sau $I \subset (-\infty, 0)$
$\displaystyle \int \frac{1}{x} dx =$ $\displaystyle \ln{ \mid x \mid } +C$
$f : \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = e^x$
$\displaystyle \int e^x dx =$ $\displaystyle e^x +C$
$f : \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = a^x$,
unde $a \gt 0$, $a \ne 1$
$\displaystyle \int a^x dx =$ $\displaystyle \frac{a^x}{ \ln {a}} +C$
$f : \mathbb{R} \rightarrow \mathbb{R}$, $\displaystyle f(x) = \frac{1}{x^2 + a^2}$,
unde $a \ne 0$
$\displaystyle \int \frac{1}{x^2 + a^2} dx =$ $\displaystyle \frac{1}{a}\text{arctg}{ \frac{x}{a} } +C$
$f : I \rightarrow \mathbb{R}$, $\displaystyle f(x) = \frac{1}{x^2 - a^2}$,
unde $I \subset \mathbb{R} \setminus \{ -a, a \}$, $a \ne 0$
$\displaystyle \int \frac{1}{x^2 - a^2} dx =$ $\displaystyle \frac{1}{2a}\ln{ \left| \frac{x-a}{x+a} \right| } +C$
$f : \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = \sin{x}$
$\displaystyle \int \sin{x} dx =$ $\displaystyle -\cos{x} +C$
$f : \mathbb{R} \rightarrow \mathbb{R}$, $f(x) = \cos{x}$
$\displaystyle \int \cos{x} dx =$ $\displaystyle \sin{x} +C$
$f : I \rightarrow \mathbb{R}$, $\displaystyle f(x) = \frac{1}{\cos^2{x}}$,
unde $\displaystyle I \subset \mathbb{R} \setminus$ $\displaystyle \left\{ \frac{ (2k+1) \pi}{2} | k \in \mathbb{Z} \right\}$
$\displaystyle \int \frac{1}{\cos^2{x}} dx =$ $\displaystyle \text{tg}{x} +C$
$f : I \rightarrow \mathbb{R}$, $\displaystyle f(x) = \frac{1}{\sin^2{x}}$,
unde $\displaystyle I \subset \mathbb{R} \setminus$ $\displaystyle \left\{ k \pi | k \in \mathbb{Z} \right\}$
$\displaystyle \int \frac{1}{\sin^2{x}} dx =$ $\displaystyle -\text{ctg}{x} +C$
$f : I \rightarrow \mathbb{R}$, $\displaystyle f(x) = \text{tg}{x}$,
unde $\displaystyle I \subset \mathbb{R} \setminus$ $\displaystyle \left\{ \frac{ (2k+1) \pi}{2} | k \in \mathbb{Z} \right\}$
$\displaystyle \int \text{tg}{x} dx =$ $\displaystyle -\ln{\left| \cos{x} \right|} +C$
$f : I \rightarrow \mathbb{R}$, $\displaystyle f(x) = \text{ctg}{x}$,
unde $\displaystyle I \subset \mathbb{R} \setminus$ $\displaystyle \left\{ k \pi | k \in \mathbb{Z} \right\}$
$\displaystyle \int \text{ctg}{x} dx =$ $\displaystyle \ln{\left| \sin{x} \right|} +C$
$f : I \rightarrow \mathbb{R}$, $\displaystyle f(x) = \frac{1}{ \sqrt{x^2 + a^2} }$,
unde $a \ne 0$
$\displaystyle \int \frac{1}{ \sqrt{x^2 + a^2} } dx =$ $\displaystyle \ln{\left( x + \sqrt{x^2+a^2} \right)} +C$
$f : I \rightarrow \mathbb{R}$, $\displaystyle f(x) = \frac{1}{ \sqrt{x^2 - a^2} }$,
unde $I \subset (-\infty, -a)$ sau $I \subset (a, \infty)$, $a \ne 0$
$\displaystyle \int \frac{1}{ \sqrt{ x^2 - a^2 } } dx =$ $\displaystyle \ln{\left| x + \sqrt{x^2-a^2} \right|} +C$
$f : I \rightarrow \mathbb{R}$, $\displaystyle f(x) = \frac{1}{ \sqrt{a^2 - x^2} }$,
unde $a \gt 0$, $I \subset (-a, a)$
$\displaystyle \int \frac{1}{ \sqrt{ a^2 - x^2 } } dx =$ $\displaystyle \arcsin{ \frac{x}{a} } +C$