integral de 0 a 1 de sqrt(1+(-1/x)^2)

Solución

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

Es divergente

Pasos de solución

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

Simplificar

1 + (- \frac{1}{x})^{2} : \frac{x ^{2} + 1}{x ^{2}}

= \int_{0}^{1} \frac{x ^{2} + 1}{x ^{2}} d x

Simplificar

\frac{x ^{2} + 1}{x ^{2}} : \frac{x ^{2} + 1}{x ^{2}}

= \int_{0}^{1} \frac{x ^{2} + 1}{x ^{2}} d x

x^{2} = (x),

asumiendo que

x \geq 0

= \int_{0}^{1} \frac{x ^{2} + 1}{x} d x

Aplicar integración por sustitución

: \int_{1}^{2} \frac{u ^{2}}{u ^{2} - 1} d u

= \int_{1}^{2} \frac{u ^{2}}{u ^{2} - 1} d u

Factorizar

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

= \int_{1}^{2} \frac{u ^{2}}{- ( - u ^{2} + 1 )} d u

Sacar la constante:

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

= - \int_{1}^{2} \frac{u ^{2}}{- u ^{2} + 1} d u

\frac{u ^{2}}{- u ^{2} + 1} = \frac{1}{- u ^{2} + 1} - 1

= - \int_{1}^{2} \frac{1}{- u ^{2} + 1} - 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

= - (\int_{1}^{2} \frac{1}{- u ^{2} + 1} d u - \int_{1}^{2} 1 d u)

\int_{1}^{2} \frac{1}{- u ^{2} + 1} d u =

Es divergente

Si parte de la expresión diverge, toda la expresión diverge

= Es divergente

\frac{( 6 )}{s ^{2} + 5 s + 6}

\int_{1}^{R} x e^{- 4 x} d x

\frac{d y}{d x} = (y - 7)^{2}

\frac{d}{d x} (x (x^{2} + 1)^{3})