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integral de 1/(a*x-x^2)

Solución

\int \frac{1}{a \cdot x - x ^{2}} d x

- \frac{1}{a} (- ln (\frac{∣ x ∣}{∣ a ∣}) + ln \frac{x}{a} - 1) + C

Pasos de solución

\int \frac{1}{a x - x ^{2}} d x

Factorizar

a x - x^{2} : - (- a x + x^{2})

= \int \frac{1}{- ( - a x + x ^{2} )} d x

Sacar la constante:

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

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

Factorizar

\frac{1}{- a x + x ^{2}}

= - \int \frac{1}{x ( - a + x )} d x

- a + x = a (- 1 + \frac{x}{a})

= - \int \frac{1}{x a ( - 1 + \frac{x}{a} )} d x

Sacar la constante:

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

= - \frac{1}{a} \cdot \int \frac{1}{x ( - 1 + \frac{x}{a} )} d x

Aplicar integración por sustitución

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

= - \frac{1}{a} \cdot \int \frac{1}{u ( u - 1 )} d u

Tomar la fracción parcial de

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

= - \frac{1}{a} \cdot \int - \frac{1}{u} + \frac{1}{u - 1} 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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= - \frac{1}{a} (- ln ∣ u ∣ + ln ∣ u - 1 ∣)

Sustituir en la ecuación

u = \frac{x}{a}

= - \frac{1}{a} (- ln \frac{x}{a} + ln \frac{x}{a} - 1)

Aplicar las propiedades de los valores absolutos:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{x}{a} = \frac{∣ x ∣}{∣ a ∣}

= - \frac{1}{a} (- ln (\frac{∣ x ∣}{∣ a ∣}) + ln \frac{x}{a} - 1)

Agregar una constante a la solución

= - \frac{1}{a} (- ln (\frac{∣ x ∣}{∣ a ∣}) + ln \frac{x}{a} - 1) + C

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