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$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C$ $(gilt fürn \neq - 1)$	$\int \frac{1}{x} d x = \ln \| x \| + C$
$\int e^{x} d x = e^{x} + C$	$\int a^{x} d x = a^{x} \cdot \frac{1}{\ln a} + C$
$\int \sin x d x = - \cos x + C$	$\int \cos x d x = \sin x + C$
$\int \frac{1}{\cos^{2} x} d x = \tan x + C$	$\int \frac{1}{\sin^{2} x} d x = - \cot x + C$
$\int \frac{1}{\sqrt{1 - x^{2}}} d x = {\begin{matrix} \arcsin x + C_{1} \\ - \arccos x + C_{2} \end{matrix}$	$\int \frac{1}{1 + x^{2}} d x = {\begin{matrix} \arctan x + C_{1} \\ - arccot x + C_{2} \end{matrix}$
$\int \sinh x d x = \cosh x + C$	$\int \cosh x d x = \sinh x + C$
$\int \frac{1}{\cosh^{2} x} d x = \tanh x + C$	$\int \frac{1}{\sinh^{2} x} d x = - \coth x + C$
$\int \frac{1}{\sqrt{x^{2} + 1}} d x = a r s i n h x + C = \ln \| x + \sqrt{x^{2} + 1} \| + C$
$\int \frac{1}{\sqrt{x^{2} - 1}} d x = a r c o s h x + C = \ln \| x + \sqrt{x^{2} - 1} \| + C (für \| x \| > 1)$
$\int \frac{1}{1 - x^{2}} d x = {\begin{matrix} a r t a n h x + C_{1} = \frac{1}{2} \cdot \ln (\frac{1 + x}{1 - x}) + C_{1} für \| x \| < 1 \\ a r c o t h x + C_{2} = \frac{1}{2} \cdot \ln (\frac{1 + x}{x - 1}) + C_{2} für \| x \| > 1 \end{matrix}$