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人気のある微分積分 >

integral of 1/x*sqrt((x-9)/(x+9))

解

\int \frac{1}{x} \cdot \frac{x - 9}{x + 9} d x

解

ln \frac{x - 9}{x + 9} + 1 - ln \frac{x - 9}{x + 9} - 1 - 2 arctan (\frac{x - 9}{x + 9}) + C

解答ステップ

\int \frac{1}{x} \frac{x - 9}{x + 9} 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}{4 ( u + 1 )} - \frac{1}{4 ( u - 1 )} - \frac{1}{2 ( u ^{2} + 1 )}

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

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

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

代用を戻す

u = \frac{x - 9}{x + 9}

= 4 (\frac{1}{4} ln \frac{x - 9}{x + 9} + 1 - \frac{1}{4} ln \frac{x - 9}{x + 9} - 1 - \frac{1}{2} arctan (\frac{x - 9}{x + 9}))

簡素化

4 (\frac{1}{4} ln \frac{x - 9}{x + 9} + 1 - \frac{1}{4} ln \frac{x - 9}{x + 9} - 1 - \frac{1}{2} arctan (\frac{x - 9}{x + 9})) : ln \frac{x - 9}{x + 9} + 1 - ln \frac{x - 9}{x + 9} - 1 - 2 arctan (\frac{x - 9}{x + 9})

= ln \frac{x - 9}{x + 9} + 1 - ln \frac{x - 9}{x + 9} - 1 - 2 arctan (\frac{x - 9}{x + 9})

定数を解答に追加する

= ln \frac{x - 9}{x + 9} + 1 - ln \frac{x - 9}{x + 9} - 1 - 2 arctan (\frac{x - 9}{x + 9}) + C

グラフ

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