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integral de 0 a 1 de 1/(x(2x+5))

Solución

\int_{0}^{1} \frac{1}{x ( 2 x + 5 )} d x

Solución

Es divergente

Pasos de solución

\int_{0}^{1} \frac{1}{x ( 2 x + 5 )} d x

Completar el cuadrado

x (2 x + 5) : 2 (x + \frac{5}{4})^{2} - \frac{25}{8}

= \int_{0}^{1} \frac{1}{2 ( x + \frac{5}{4} ) ^{2} - \frac{25}{8}} d x

Aplicar integración por sustitución

: \int_{\frac{5}{4}}^{\frac{9}{4}} \frac{8}{16 u ^{2} - 25} d u

= \int_{\frac{5}{4}}^{\frac{9}{4}} \frac{8}{16 u ^{2} - 25} d u

Sacar la constante:

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

= 8 \cdot \int_{\frac{5}{4}}^{\frac{9}{4}} \frac{1}{16 u ^{2} - 25} d u

Aplicar integración por sustitución

= 8 \cdot \int_{1}^{\frac{9}{5}} \frac{1}{20 ( v ^{2} - 1 )} d v

Sacar la constante:

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

= 8 \cdot \frac{1}{20} \cdot \int_{1}^{\frac{9}{5}} \frac{1}{v ^{2} - 1} d v

Factorizar

v^{2} - 1 : - (- v^{2} + 1)

= 8 \cdot \frac{1}{20} \cdot \int_{1}^{\frac{9}{5}} \frac{1}{- ( - v ^{2} + 1 )} d v

Sacar la constante:

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

= 8 \cdot \frac{1}{20} (- \int_{1}^{\frac{9}{5}} \frac{1}{- v ^{2} + 1} d v)

Aplicar la regla de integración:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 8 \cdot \frac{1}{20} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{1}^{\frac{9}{5}})

Simplificar

8 \cdot \frac{1}{20} (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{1}^{\frac{9}{5}}) : - \frac{2}{5} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{1}^{\frac{9}{5}}

= - \frac{2}{5} [\frac{ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣}{2}]_{1}^{\frac{9}{5}}

Calcular los limites:

Es divergente

= Es divergente

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