integral of 1/(e^{2x)+1}

Lösung

\int \frac{1}{e ^{2 x} + 1} d x

- \frac{1}{2} ln e^{2 x} + 1 + x + C

Schritte zur Lösung

\int \frac{1}{e ^{2 x} + 1} d x

\frac{1}{e ^{2 x} + 1} = - \frac{e ^{2 x}}{e ^{2 x} + 1} + 1

= \int - \frac{e ^{2 x}}{e ^{2 x} + 1} + 1 d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{e ^{2 x}}{e ^{2 x} + 1} d x + \int 1 d x

\int \frac{e ^{2 x}}{e ^{2 x} + 1} d x = \frac{1}{2} ln e^{2 x} + 1

\int 1 d x = x

= - \frac{1}{2} ln e^{2 x} + 1 + x

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2} ln e^{2 x} + 1 + x + C

\int (4 - x) x d x

y^{^{'}} + 3 y = t + e^{(- 2 t)}

x \to - 3 lim (\frac{1}{( x + 3 ) ^{2}})

y^{^{'}} - e^{x} e^{y} = - 1

\frac{\partial}{\partial x} (e^{1 + x^{2} - y^{2}})