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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

定数を除く:

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

= \frac{1}{2} \cdot \int \frac{1}{u ( u - 1 )} d u

u置換積分法を適用する

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

因数

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

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

共通積分を使用する:

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

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

代用を戻す

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of+3x^5sqrt(x^3+1)

\int + 3 x^{5} x^{3} + 1 d x

integral of 1/(tan(x/5))

\int \frac{1}{tan ( \frac{x}{5} )} d x

integral of+1/(sqrt(12-4x-x^2))

\int + \frac{1}{12 - 4 x - x ^{2}} d x

integral of 3^{5x+1}

\int 3^{5 x + 1} d x

integral of [(2x-3)/(x^2+2x+5)]

\int [\frac{2 x - 3}{x ^{2} + 2 x + 5}] d x