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人気のある微分積分 >

integral of sqrt((1-x)/(1+x))*1/x

解

\int \frac{1 - x}{1 + x} \cdot \frac{1}{x} d x

解

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

解答ステップ

\int \frac{1 - x}{1 + x} \frac{1}{x} d x

u置換積分法を適用する

= \int \frac{4 u ^{2}}{( u ^{2} + 1 ) ( u ^{2} - 1 )} d u

定数を除く:

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

= 4 \cdot \int \frac{u ^{2}}{( u ^{2} + 1 ) ( u ^{2} - 1 )} d u

以下の部分分数を得る:

\frac{u ^{2}}{( u ^{2} + 1 ) ( u ^{2} - 1 )} : \frac{1}{2 ( u ^{2} + 1 )} - \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u - 1 )}

= 4 \cdot \int \frac{1}{2 ( u ^{2} + 1 )} - \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u - 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

= 4 (\int \frac{1}{2 ( u ^{2} + 1 )} d u - \int \frac{1}{4 ( u + 1 )} d u + \int \frac{1}{4 ( u - 1 )} d u)

\int \frac{1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} arctan (u)

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

= 4 (\frac{1}{2} arctan (u) - \frac{1}{4} ln ∣ u + 1 ∣ + \frac{1}{4} ln ∣ u - 1 ∣)

代用を戻す

u = \frac{1 - x}{1 + x}

= 4 (\frac{1}{2} arctan (\frac{1 - x}{1 + x}) - \frac{1}{4} ln \frac{1 - x}{1 + x} + 1 + \frac{1}{4} ln \frac{1 - x}{1 + x} - 1)

簡素化

4 (\frac{1}{2} arctan (\frac{1 - x}{1 + x}) - \frac{1}{4} ln \frac{1 - x}{1 + x} + 1 + \frac{1}{4} ln \frac{1 - x}{1 + x} - 1) : 2 arctan (\frac{1 - x}{1 + x}) - ln \frac{1 - x}{1 + x} + 1 + ln \frac{1 - x}{1 + x} - 1

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

定数を解答に追加する

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

グラフ

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