inverselaplace arctan(s)+1

Lösung

inverse laplace transformation

arctan (s) + 1

- \frac{sin ( t )}{t}

Schritte zur Lösung

L^{- 1} {arctan (s) + 1}

Verwende inverse Laplace-Transformationstabelle:

L^{- 1} {F (s)} = - \frac{1}{t} L^{- 1} {F^{'} (s)}

L^{- 1} {arctan (s) + 1} = - \frac{1}{t} L^{- 1} {(arctan (s) + 1)^{'}}

= - \frac{1}{t} L^{- 1} {(arctan (s) + 1)^{^{'}}}

(arctan (s) + 1)^{'} = \frac{1}{s ^{2} + 1}

= - \frac{1}{t} L^{- 1} {\frac{1}{s ^{2} + 1}}

L^{- 1} {\frac{1}{s ^{2} + 1}} : sin (t)

= - \frac{1}{t} sin (t)

= - \frac{sin ( t )}{t}

\int \frac{cos ( x ) - \frac{s i n ( x )}{x}}{x} d x

\frac{\partial}{\partial x} (- 2 x y (6 - x - y))

x \to 0 lim (lo g_{3} (x))

d er i v a t i v e 4 x + 1

\int \frac{2 x - 3}{( x ^{2} - 3 x + 6 ) ^{2}} d x