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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

置換積分法を適用する

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

\frac{1}{u ( a ^{2} - a ^{2} u ^{2} )} = \frac{1}{a ^{2}} \frac{1}{u ( 1 - u ^{2} )}

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

定数を除く:

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

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

u置換積分法を適用する

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

定数を除く:

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

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

以下の部分分数を得る:

\frac{1}{v ( 1 - v )} : \frac{1}{v} - \frac{1}{v - 1}

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

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

\int \frac{1}{v} d v = ln ∣ v ∣

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

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

代用を戻す

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of 1/(4+9x^2)

\int \frac{1}{4 + 9 x ^{2}} d x

勾配 y=8x-x^2(1.7)

s l o p e y = 8 x - x^{2} (1.7)

limit as x approaches 0 of e^x+1/x

x \to 0 lim (e^{x} + \frac{1}{x})

derivative of 0.8x^{1.3}

\frac{d}{d x} (0.8 x^{1.3})

limit as x approaches 3-of-x+5

x \to 3 - lim (- x + 5)