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integral de (1.05)/(1+100e^{-0.285x)}

Solución

\int \frac{1.05}{1 + 100 e ^{- 0.285 x}} d x

- 3.68421 \dots (- 0.285 x - ln 100 e^{- 0.285 x} + 1) + C

Pasos de solución

\int \frac{1.05}{1 + 100 e ^{- 0.285 x}} d x

Sacar la constante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 1.05 \cdot \int \frac{1}{1 + 100 e ^{- 0.285 x}} d x

\frac{1}{1 + 100 e ^{- 0.285 x}} = - \frac{100 e ^{- 0.285 x}}{1 + 100 e ^{- 0.285 x}} + 1

= 1.05 \cdot \int - \frac{100 e ^{- 0.285 x}}{1 + 100 e ^{- 0.285 x}} + 1 d x

Aplicar integración por sustitución

: \int - \frac{1}{0.285 ( 1 + 100 e ^{u} )} d u

= 1.05 \cdot \int - \frac{1}{0.285 ( 1 + 100 e ^{u} )} d u

Sacar la constante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 1.05 (- \frac{1}{0.285} \cdot \int \frac{1}{1 + 100 e ^{u}} d u)

Simplificar

= 1.05 (- 3.50877 \dots \cdot \int \frac{1}{1 + 100 e ^{u}} d u)

Aplicar integración por sustitución

: \int \frac{1}{v ( 1 + 100 v )} d v

= 1.05 (- 3.50877 \dots \cdot \int \frac{1}{v ( 1 + 100 v )} d v)

Tomar la fracción parcial de

\frac{1}{v ( 1 + 100 v )} : \frac{1}{v} - \frac{100}{100 v + 1}

= 1.05 (- 3.50877 \dots \cdot \int \frac{1}{v} - \frac{100}{100 v + 1} d v)

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

= 1.05 (- 3.50877 \dots (\int \frac{1}{v} d v - \int \frac{100}{100 v + 1} d v))

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{100}{100 v + 1} d v = ln ∣ 100 v + 1 ∣

= 1.05 (- 3.50877 \dots (ln ∣ v ∣ - ln ∣ 100 v + 1 ∣))

Sustitución hacia atrás

= 1.05 (- 3.50877 \dots (ln e^{- 0.285 x} - ln 100 e^{- 0.285 x} + 1))

Simplificar

1.05 (- 3.50877 \dots (ln e^{- 0.285 x} - ln 100 e^{- 0.285 x} + 1)) : - 3.68421 \dots (- 0.285 x - ln 100 e^{- 0.285 x} + 1)

= - 3.68421 \dots (- 0.285 x - ln 100 e^{- 0.285 x} + 1)

Agregar una constante a la solución

= - 3.68421 \dots (- 0.285 x - ln 100 e^{- 0.285 x} + 1) + C

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