integral of 1/(1+e^{6x)}

Lösung

\int \frac{1}{1 + e ^{6 x}} d x

- \frac{1}{6} ln 1 + e^{6 x} + x + C

Schritte zur Lösung

\int \frac{1}{1 + e ^{6 x}} d x

\frac{1}{1 + e ^{6 x}} = - \frac{e ^{6 x}}{1 + e ^{6 x}} + 1

= \int - \frac{e ^{6 x}}{1 + e ^{6 x}} + 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 ^{6 x}}{1 + e ^{6 x}} d x + \int 1 d x

\int \frac{e ^{6 x}}{1 + e ^{6 x}} d x = \frac{1}{6} ln 1 + e^{6 x}

\int 1 d x = x

= - \frac{1}{6} ln 1 + e^{6 x} + x

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{6} ln 1 + e^{6 x} + x + C

n = 1 \sum \infty \frac{5}{n + 2}

\int (x^{3} + 3) d x

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

a re a cos (x), sin (x)

2 x^{2} y^{^{''}} + 3 x y^{^{'}} = y