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人気のある微分積分 >

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

解

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

解

- \frac{1}{16} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2}) + C

解答ステップ

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

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 + x ^{2}}{x} d x

u置換積分法を適用する

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

因数

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

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

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

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

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

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

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

\int 1 d u = u

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

代用を戻す

u = 1 + x^{2}

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

簡素化

\frac{1}{8} (- (\frac{ln 1 + x ^{2} + 1}{2} - \frac{ln 1 + x ^{2} - 1}{2} - 1 + x^{2})) : - \frac{1}{16} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2})

= - \frac{1}{16} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2})

定数を解答に追加する

= - \frac{1}{16} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2}) + C

グラフ

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