(ln(sqrt((1+x)/(1-x))))^'

Lösung

(ln (\frac{1 + x}{1 - x}))^{^{'}}

\frac{1}{( - x + 1 ) ( x + 1 )}

Schritte zur Lösung

(ln (\frac{1 + x}{1 - x}))^{'}

Wende die Kettenregel an:

\frac{1}{\frac{1 + x}{1 - x}} (\frac{1 + x}{1 - x})^{'}

= \frac{1}{\frac{1 + x}{1 - x}} (\frac{1 + x}{1 - x})^{'}

(\frac{1 + x}{1 - x})^{'} = \frac{1}{x + 1 ( - x + 1 ) ^{\frac{3}{2}}}

= \frac{1}{\frac{1 + x}{1 - x}} \cdot \frac{1}{x + 1 ( - x + 1 ) ^{\frac{3}{2}}}

Vereinfache

\frac{1}{\frac{1 + x}{1 - x}} \cdot \frac{1}{x + 1 ( - x + 1 ) ^{\frac{3}{2}}} : \frac{1}{( - x + 1 ) ( x + 1 )}

= \frac{1}{( - x + 1 ) ( x + 1 )}

\int \frac{( x ^{2} - 1 )}{x ^{2} + 1} d x

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