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人気のある微分積分 >

integral of sqrt((x-2)/(x+2))

解

\int \frac{x - 2}{x + 2} d x

解

8 - \frac{1}{4} ln \frac{x - 2}{x + 2} + 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} + 1 )} + \frac{1}{4} ln \frac{x - 2}{x + 2} - 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} - 1 )} + C

解答ステップ

\int \frac{x - 2}{x + 2} d x

u置換積分法を適用する

= \int \frac{8 u ^{2}}{- u ^{2} + 1 ( - u ^{2} + 1 ) ^{\frac{3}{2}}} d u

定数を除く:

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

= 8 \cdot \int \frac{u ^{2}}{- u ^{2} + 1 ( - u ^{2} + 1 ) ^{\frac{3}{2}}} d u

(- u^{2} + 1)^{\frac{3}{2}} = (- u^{2} + 1) - u^{2} + 1

= 8 \cdot \int \frac{u ^{2}}{- u ^{2} + 1 ( - u ^{2} + 1 ) - u ^{2} + 1} d u

- u^{2} + 1 (- u^{2} + 1) - u^{2} + 1 = (- u^{2} + 1)^{2}

= 8 \cdot \int \frac{u ^{2}}{( - u ^{2} + 1 ) ^{2}} d u

以下の部分分数を得る:

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

= 8 \cdot \int - \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u + 1 ) ^{2}} + \frac{1}{4 ( u - 1 )} + \frac{1}{4 ( u - 1 ) ^{2}} 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

= 8 (- \int \frac{1}{4 ( u + 1 )} d u + \int \frac{1}{4 ( u + 1 ) ^{2}} d u + \int \frac{1}{4 ( u - 1 )} d u + \int \frac{1}{4 ( u - 1 ) ^{2}} d u)

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

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

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

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

= 8 (- \frac{1}{4} ln ∣ u + 1 ∣ - \frac{1}{4 ( u + 1 )} + \frac{1}{4} ln ∣ u - 1 ∣ - \frac{1}{4 ( u - 1 )})

代用を戻す

u = \frac{x - 2}{x + 2}

= 8 - \frac{1}{4} ln \frac{x - 2}{x + 2} + 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} + 1 )} + \frac{1}{4} ln \frac{x - 2}{x + 2} - 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} - 1 )}

定数を解答に追加する

= 8 - \frac{1}{4} ln \frac{x - 2}{x + 2} + 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} + 1 )} + \frac{1}{4} ln \frac{x - 2}{x + 2} - 1 - \frac{1}{4 ( \frac{x - 2}{x + 2} - 1 )} + C

グラフ

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