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人気のある微分積分 >

integral of ((1+x^2)^{1/2})/x

解

\int \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{x} d x

解

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

解答ステップ

\int \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{x} d x

u置換積分法を適用する

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

因数

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

= \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

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

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

= - \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

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

代用を戻す

u = (1 + x^{2})^{\frac{1}{2}}

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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