integral de 1/(e^x-9e^{-x)}

Solución

\int \frac{1}{e ^{x} - 9 e ^{- x}} d x

- \frac{1}{6} ln ∣ e^{x} + 3 ∣ + \frac{1}{6} ln ∣ e^{x} - 3 ∣ + C

Pasos de solución

\int \frac{1}{e ^{x} - 9 e ^{- x}} d x

Factorizar

e^{x} - 9 e^{- x} : e^{- x} (e^{x} + 3) (e^{x} - 3)

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

Aplicar las leyes de los exponentes:

\frac{1}{a ^{b}} = a^{- b}

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

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

Simplificar

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

= \int \frac{e ^{x}}{( e ^{x} + 3 ) ( e ^{x} - 3 )} d x

Aplicar integración por sustitución

: \int \frac{1}{( u + 3 ) ( u - 3 )} d u

= \int \frac{1}{( u + 3 ) ( u - 3 )} d u

Tomar la fracción parcial de

\frac{1}{( u + 3 ) ( u - 3 )} : - \frac{1}{6 ( u + 3 )} + \frac{1}{6 ( u - 3 )}

= \int - \frac{1}{6 ( u + 3 )} + \frac{1}{6 ( u - 3 )} d u

Aplicar la regla de la suma:

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

= - \int \frac{1}{6 ( u + 3 )} d u + \int \frac{1}{6 ( u - 3 )} d u

\int \frac{1}{6 ( u + 3 )} d u = \frac{1}{6} ln ∣ u + 3 ∣

\int \frac{1}{6 ( u - 3 )} d u = \frac{1}{6} ln ∣ u - 3 ∣

= - \frac{1}{6} ln ∣ u + 3 ∣ + \frac{1}{6} ln ∣ u - 3 ∣

Sustituir en la ecuación

u = e^{x}

= - \frac{1}{6} ln ∣ e^{x} + 3 ∣ + \frac{1}{6} ln ∣ e^{x} - 3 ∣

Agregar una constante a la solución

= - \frac{1}{6} ln ∣ e^{x} + 3 ∣ + \frac{1}{6} ln ∣ e^{x} - 3 ∣ + C

x \to \infty + lim (tan (x))

x \to 0 lim (\frac{arctan ( x )}{x})

\frac{d}{d x} (\frac{8 x ^{2} + 8 x + 4}{x})

\frac{\partial}{\partial x} (x + 2 y)

\int \frac{5 x + 4}{( x + 8 ) ^{2} ( x - 1 )} d x