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人気のある微分積分 >

integral of (sqrt(64+x^2))/x

解

\int \frac{64 + x ^{2}}{x} d x

解

- 4 ln \frac{1}{8} x^{2} + 64 + 1 + 4 ln \frac{1}{8} x^{2} + 64 - 1 + 64 + x^{2} + C

解答ステップ

\int \frac{64 + x ^{2}}{x} d x

u置換積分法を適用する

= \int \frac{u ^{2}}{u ^{2} - 64} d u

因数

u^{2} - 64 : - (- u^{2} + 64)

= \int \frac{u ^{2}}{- ( - u ^{2} + 64 )} d u

定数を除く:

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

= - \int \frac{u ^{2}}{- u ^{2} + 64} d u

置換積分法を適用する

= - \int \frac{8 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

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

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

= - 8 \cdot \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

= - 8 (\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

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

代用を戻す

= - 8 \frac{ln \frac{64 + x ^{2}}{8} + 1}{2} - \frac{ln \frac{64 + x ^{2}}{8} - 1}{2} - \frac{64 + x ^{2}}{8}

簡素化

- 8 \frac{ln \frac{64 + x ^{2}}{8} + 1}{2} - \frac{ln \frac{64 + x ^{2}}{8} - 1}{2} - \frac{64 + x ^{2}}{8} : - 4 ln \frac{1}{8} x^{2} + 64 + 1 + 4 ln \frac{1}{8} x^{2} + 64 - 1 + 64 + x^{2}

= - 4 ln \frac{1}{8} x^{2} + 64 + 1 + 4 ln \frac{1}{8} x^{2} + 64 - 1 + 64 + x^{2}

定数を解答に追加する

= - 4 ln \frac{1}{8} x^{2} + 64 + 1 + 4 ln \frac{1}{8} x^{2} + 64 - 1 + 64 + x^{2} + C

グラフ

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