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人気のある微分積分 >

integral of ((2x-3))/((x-1)^2(x-2)^2)

解

\int \frac{( 2 x - 3 )}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

解

- \frac{1}{x - 2} + \frac{1}{x - 1} + C

解答ステップ

\int \frac{2 x - 3}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

u置換積分法を適用する

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

拡張

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

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

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

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

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

代用を戻す

u = x - 1

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

簡素化

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

= - \frac{1}{x - 2} + \frac{1}{x - 1}

定数を解答に追加する

= - \frac{1}{x - 2} + \frac{1}{x - 1} + C

グラフ

人気の例

integral of e^{-5/6 x}

\int e^{- \frac{5}{6} x} d x

integral of vsin(v)

\int v sin (v) d v

integral of 16e^{4x}

\int 16 e^{4 x} d x

integral of (8x^3-12x^2+7x+2)/(2x-2)

\int \frac{8 x ^{3} - 12 x ^{2} + 7 x + 2}{2 x - 2} d x

integral of 6/x+5/(x^2)

\int \frac{6}{x} + \frac{5}{x ^{2}} d x