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人気のある微分積分 >

integral of (sqrt(1+x^8))/x

解

\int \frac{1 + x ^{8}}{x} d x

解

- \frac{1}{8} (ln x^{8} + 1 + 1 - ln x^{8} + 1 - 1 - 2 x^{8} + 1) + C

解答ステップ

\int \frac{1 + x ^{8}}{x} d x

u置換積分法を適用する

= \int \frac{1 + u}{8 u} d u

定数を除く:

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

= \frac{1}{8} \cdot \int \frac{1 + u}{u} d u

u置換積分法を適用する

= \frac{1}{8} \cdot \int \frac{2 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

= \frac{1}{8} \cdot 2 \cdot \int \frac{v ^{2}}{v ^{2} - 1} d v

因数

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

= \frac{1}{8} \cdot 2 \cdot \int \frac{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

= \frac{1}{8} \cdot 2 (- \int \frac{v ^{2}}{- v ^{2} + 1} d v)

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

= \frac{1}{8} \cdot 2 (- \int \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

= \frac{1}{8} \cdot 2 (- (\int \frac{1}{- v ^{2} + 1} d v - \int 1 d v))

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

\int 1 d v = v

= \frac{1}{8} \cdot 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v))

代用を戻す

= \frac{1}{8} \cdot 2 (- (\frac{ln 1 + x ^{8} + 1}{2} - \frac{ln 1 + x ^{8} - 1}{2} - 1 + x^{8}))

簡素化

\frac{1}{8} \cdot 2 (- (\frac{ln 1 + x ^{8} + 1}{2} - \frac{ln 1 + x ^{8} - 1}{2} - 1 + x^{8})) : - \frac{1}{8} (ln x^{8} + 1 + 1 - ln x^{8} + 1 - 1 - 2 x^{8} + 1)

= - \frac{1}{8} (ln x^{8} + 1 + 1 - ln x^{8} + 1 - 1 - 2 x^{8} + 1)

定数を解答に追加する

= - \frac{1}{8} (ln x^{8} + 1 + 1 - ln x^{8} + 1 - 1 - 2 x^{8} + 1) + C

グラフ

人気の例

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