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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

7.40474 \dots

解答ステップ

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

結合

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

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

簡素化

\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), （ x \geq 0

と仮定

）

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

u置換積分法を適用する

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

因数

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

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

定数を除く:

\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

総和規則を適用する:

\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))

簡素化

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

グラフ

人気の例

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

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\int_{0}^{1} x π^{x} d x

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

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