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人気のある微分積分 >

integral from 1 to 5 of (sqrt(24+x^2))/x

解

\int_{1}^{5} \frac{24 + x ^{2}}{x} d x

解

- 6 ln (\frac{217 - 88 6}{25}) + 2

+1

十進法表記

8.98308 \dots

解答ステップ

\int_{1}^{5} \frac{24 + x ^{2}}{x} d x

u置換積分法を適用する

= \int_{5}^{7} \frac{u ^{2}}{u ^{2} - 24} d u

因数

u^{2} - 24 : - (- u^{2} + 24)

= \int_{5}^{7} \frac{u ^{2}}{- ( - u ^{2} + 24 )} d u

定数を除く:

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

= - \int_{5}^{7} \frac{u ^{2}}{- u ^{2} + 24} d u

置換積分法を適用する

= - \int_{\frac{5}{2 6}}^{\frac{7}{2 6}} \frac{2 6 v ^{2}}{- v ^{2} + 1} d v

定数を除く:

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

= - 26 \cdot \int_{\frac{5}{2 6}}^{\frac{7}{2 6}} \frac{v ^{2}}{- v ^{2} + 1} d v

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

= - 26 \cdot \int_{\frac{5}{2 6}}^{\frac{7}{2 6}} \frac{1}{- v ^{2} + 1} - 1 d v

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

= - 26 (\int_{\frac{5}{2 6}}^{\frac{7}{2 6}} \frac{1}{- v ^{2} + 1} d v - \int_{\frac{5}{2 6}}^{\frac{7}{2 6}} 1 d v)

\int_{\frac{5}{2 6}}^{\frac{7}{2 6}} \frac{1}{- v ^{2} + 1} d v = \frac{ln ( \frac{217 - 88 6}{25} )}{2}

\int_{\frac{5}{2 6}}^{\frac{7}{2 6}} 1 d v = \frac{1}{6}

= - 26 \frac{ln ( \frac{217 - 88 6}{25} )}{2} - \frac{1}{6}

簡素化

- 26 \frac{ln ( \frac{217 - 88 6}{25} )}{2} - \frac{1}{6} : - 6 ln (\frac{217 - 88 6}{25}) + 2

= - 6 ln (\frac{217 - 88 6}{25}) + 2

グラフ

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