de

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Beliebt Rechnen >

integral from 1 to 8 of sqrt(1+1/(x^2))

Lösung

\int_{1}^{8} 1 + \frac{1}{x ^{2}} d x

Lösung

- \frac{ln ( 65 + 1 ) - ln ( ( 65 - 1 ) ( 2 + 1 ) ) + ln ( 2 - 1 )}{2} + 65 - 2

+1

Dezimale

7.40474 \dots

Schritte zur Lösung

\int_{1}^{8} 1 + \frac{1}{x ^{2}} d x

Füge

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

zusammen:

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

= \int_{1}^{8} \frac{x ^{2} + 1}{x ^{2}} d x

Vereinfache

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

= \int_{1}^{8} \frac{x ^{2} + 1}{x ^{2}} d x

x^{2} = (x),

angenommen

x \geq 0

= \int_{1}^{8} \frac{x ^{2} + 1}{x} d x

Wende U-Substitution an

= \int_{2}^{65} \frac{u ^{2}}{u ^{2} - 1} d u

Faktorisiere

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

= \int_{2}^{65} \frac{u ^{2}}{- ( - u ^{2} + 1 )} d u

Entferne die Konstante:

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

= - \int_{2}^{65} \frac{u ^{2}}{- u ^{2} + 1} d u

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

= - \int_{2}^{65} \frac{1}{- u ^{2} + 1} - 1 d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int_{2}^{65} \frac{1}{- u ^{2} + 1} d u - \int_{2}^{65} 1 d u)

\int_{2}^{65} \frac{1}{- u ^{2} + 1} d u = \frac{ln ( 65 + 1 ) - ln ( ( 65 - 1 ) ( 2 + 1 ) ) + ln ( 2 - 1 )}{2}

\int_{2}^{65} 1 d u = 65 - 2

= - (\frac{ln ( 65 + 1 ) - ln ( ( 65 - 1 ) ( 2 + 1 ) ) + ln ( 2 - 1 )}{2} - (65 - 2))

Vereinfache

= - \frac{ln ( 65 + 1 ) - ln ( ( 65 - 1 ) ( 2 + 1 ) ) + ln ( 2 - 1 )}{2} + 65 - 2

Graph

Beliebte Beispiele

integral from 1 to 2 of 1/(1+x^4)

\int_{1}^{2} \frac{1}{1 + x ^{4}} d x

integral from 0 to 1 of 1/(sqrt(-ln(x)))

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

integral from 0 to 1 of xpi^x

\int_{0}^{1} x π^{x} d x

integral from 0 to ln(15) of tanh(x)

\int_{0}^{l n (15)} tanh (x) d x

integral from-1 to 3 of (12x^3-2x+5)

\int_{- 1}^{3} (12 x^{3} - 2 x + 5) d x