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人気のある微分積分 >

integral of sqrt((1-x^2)/(x^2))

解

\int \frac{1 - x ^{2}}{x ^{2}} d x

解

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

解答ステップ

\int \frac{1 - x ^{2}}{x ^{2}} d x

簡素化

\frac{1 - x ^{2}}{x ^{2}} : \frac{1 - x ^{2}}{x}

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

u置換積分法を適用する

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of sqrt(4x^2-256)

\int 4 x^{2} - 256 d x

integral of (x-1)/(x^2-2x-3)

\int \frac{x - 1}{x ^{2} - 2 x - 3} d x

integral of 1/(y*ln(y))

\int \frac{1}{y \cdot ln ( y )} d y

integral of (130)/((x+4)^2)

\int \frac{130}{( x + 4 ) ^{2}} d x

integral of 6x^2(2x^3-1)^{99}

\int 6 x^{2} (2 x^{3} - 1)^{99} d x