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人気のある微分積分 >

integral of (22x^3-44x^2+2)/(x^2-2x)

解

\int \frac{22 x ^{3} - 44 x ^{2} + 2}{x ^{2} - 2 x} d x

解

11 x^{2} + ln ∣ x - 2 ∣ - ln ∣ x ∣ - 88 + C

解答ステップ

\int \frac{22 x ^{3} - 44 x ^{2} + 2}{x ^{2} - 2 x} d x

平方完成する

x^{2} - 2 x : (x - 1)^{2} - 1

= \int \frac{22 x ^{3} - 44 x ^{2} + 2}{( x - 1 ) ^{2} - 1} d x

u置換積分法を適用する

= \int \frac{22 ( u + 1 ) ^{3} - 44 ( u + 1 ) ^{2} + 2}{u ^{2} - 1} d u

拡張

\frac{22 ( u + 1 ) ^{3} - 44 ( u + 1 ) ^{2} + 2}{u ^{2} - 1} : \frac{22 ( u + 1 ) ^{3}}{u ^{2} - 1} - \frac{44 ( u + 1 ) ^{2}}{u ^{2} - 1} + \frac{2}{u ^{2} - 1}

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

= \int \frac{22 ( u + 1 ) ^{3}}{u ^{2} - 1} d u - \int \frac{44 ( u + 1 ) ^{2}}{u ^{2} - 1} d u + \int \frac{2}{u ^{2} - 1} d u

\int \frac{22 ( u + 1 ) ^{3}}{u ^{2} - 1} d u = 22 (\frac{1}{2} (u - 1)^{2} + 4 (u - 1) + 4 ln ∣ u - 1 ∣)

\int \frac{44 ( u + 1 ) ^{2}}{u ^{2} - 1} d u = 44 (2 ln ∣ u - 1 ∣ + u)

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

= 22 (\frac{1}{2} (u - 1)^{2} + 4 (u - 1) + 4 ln ∣ u - 1 ∣) - 44 (2 ln ∣ u - 1 ∣ + u) - ln ∣ u + 1 ∣ + ln ∣ u - 1 ∣

代用を戻す

u = x - 1

= 22 (\frac{1}{2} (x - 1 - 1)^{2} + 4 (x - 1 - 1) + 4 ln ∣ x - 1 - 1 ∣) - 44 (2 ln ∣ x - 1 - 1 ∣ + x - 1) - ln ∣ x - 1 + 1 ∣ + ln ∣ x - 1 - 1 ∣

簡素化

22 (\frac{1}{2} (x - 1 - 1)^{2} + 4 (x - 1 - 1) + 4 ln ∣ x - 1 - 1 ∣) - 44 (2 ln ∣ x - 1 - 1 ∣ + x - 1) - ln ∣ x - 1 + 1 ∣ + ln ∣ x - 1 - 1 ∣ : 11 x^{2} + ln ∣ x - 2 ∣ - ln ∣ x ∣ - 88

= 11 x^{2} + ln ∣ x - 2 ∣ - ln ∣ x ∣ - 88

定数を解答に追加する

= 11 x^{2} + ln ∣ x - 2 ∣ - ln ∣ x ∣ - 88 + C

グラフ

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