(e^xdx)/(-(e^x+1)^3)=(e^ydy)/((e^y+1)^2)

Lösung

\frac{e ^{x} d x}{- ( e ^{x} + 1 ) ^{3}} = \frac{e ^{y} d y}{( e ^{y} + 1 ) ^{2}}

y = ln (\frac{2 ( e ^{x} + 1 ) ^{2}}{- 1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} - \frac{2 c _{1} ( e ^{x} + 1 ) ^{2}}{- 1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} + \frac{1}{- 1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}})

Schritte zur Lösung

\frac{e ^{x} d x}{- ( e ^{x} + 1 ) ^{3}} = \frac{e ^{y} d y}{( e ^{y} + 1 ) ^{2}}

Solve separable ODE:

y = ln (\frac{2 ( e ^{x} + 1 ) ^{2}}{- 1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} - \frac{2 c _{1} ( e ^{x} + 1 ) ^{2}}{- 1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} + \frac{1}{- 1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}})

y = ln (\frac{2 ( e ^{x} + 1 ) ^{2}}{- 1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} - \frac{2 c _{1} ( e ^{x} + 1 ) ^{2}}{- 1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}} + \frac{1}{- 1 + 2 c _{1} ( e ^{x} + 1 ) ^{2}})

\int sec (5 x) d x

in v erse l a pl a ce \frac{1}{s ( s ^{2} - 2 s + 3 )}

d er i v a t i v e ln (\frac{1}{x ^{2}})

l a pl a ce t r an s f or m \frac{5}{3}

a re a x^{2}, 2 x - 1, 0