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人気のある微分積分 >

integral from 2 to 3 of (sqrt(x^2+1))/x

解

\int_{2}^{3} \frac{x ^{2} + 1}{x} d x

解

- \frac{ln ( 10 + 1 ) - ln ( 5 2 + 10 - 5 - 1 ) + ln ( 5 - 1 )}{2} + 10 - 5

+1

十進法表記

1.07997 \dots

解答ステップ

\int_{2}^{3} \frac{x ^{2} + 1}{x} d x

u置換積分法を適用する

= \int_{5}^{10} \frac{u ^{2}}{u ^{2} - 1} d u

因数

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

= \int_{5}^{10} \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_{5}^{10} \frac{u ^{2}}{- u ^{2} + 1} d u

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

= - \int_{5}^{10} \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_{5}^{10} \frac{1}{- u ^{2} + 1} d u - \int_{5}^{10} 1 d u)

\int_{5}^{10} \frac{1}{- u ^{2} + 1} d u = \frac{ln ( 10 + 1 ) - ln ( 5 2 + 10 - 5 - 1 ) + ln ( 5 - 1 )}{2}

\int_{5}^{10} 1 d u = 10 - 5

= - (\frac{ln ( 10 + 1 ) - ln ( 5 2 + 10 - 5 - 1 ) + ln ( 5 - 1 )}{2} - (10 - 5))

簡素化

= - \frac{ln ( 10 + 1 ) - ln ( 5 2 + 10 - 5 - 1 ) + ln ( 5 - 1 )}{2} + 10 - 5

グラフ

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