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integral de 1/(x(x+0.5))

Solución

\int \frac{1}{x ( x + 0.5 )} d x

2 (- ln (2 ∣ x + 0.5 ∣) + ln \frac{1}{0.5} (x + 0.5) - 1) + C

Pasos de solución

\int \frac{1}{x ( x + 0.5 )} d x

Aplicar integración por sustitución

: \int \frac{1}{( u - 0.5 ) u} d u

= \int \frac{1}{( u - 0.5 ) u} d u

u - 0.5 = 0.5 (\frac{u}{0.5} - 1)

= \int \frac{1}{0.5 ( \frac{u}{0.5} - 1 ) u} d u

Sacar la constante:

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

= \frac{1}{0.5} \cdot \int \frac{1}{( \frac{u}{0.5} - 1 ) u} d u

Simplificar

= 2 \cdot \int \frac{1}{( \frac{u}{0.5} - 1 ) u} d u

Aplicar integración por sustitución

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

= 2 \cdot \int \frac{1}{v ( v - 1 )} d v

Tomar la fracción parcial de

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

= 2 \cdot \int - \frac{1}{v} + \frac{1}{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

= 2 (- \int \frac{1}{v} d v + \int \frac{1}{v - 1} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= 2 (- ln ∣ v ∣ + ln ∣ v - 1 ∣)

Sustitución hacia atrás

= 2 (- ln \frac{x + 0.5}{0.5} + ln \frac{x + 0.5}{0.5} - 1)

Simplificar

2 (- ln \frac{x + 0.5}{0.5} + ln \frac{x + 0.5}{0.5} - 1) : 2 (- ln (2 ∣ x + 0.5 ∣) + ln \frac{1}{0.5} (x + 0.5) - 1)

= 2 (- ln (2 ∣ x + 0.5 ∣) + ln \frac{1}{0.5} (x + 0.5) - 1)

Agregar una constante a la solución

= 2 (- ln (2 ∣ x + 0.5 ∣) + ln \frac{1}{0.5} (x + 0.5) - 1) + C

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