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人気のある微分積分 >

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

解

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

解

2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{5}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) + C

解答ステップ

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

平方完成する

x^{2} - 5 x + 6 : (x - \frac{5}{2})^{2} - \frac{1}{4}

= \int \frac{x}{( x - \frac{5}{2} ) ^{2} - \frac{1}{4}} d x

u置換積分法を適用する

= \int \frac{2 ( 2 u + 5 )}{4 u ^{2} - 1} d u

定数を除く:

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

= 2 \cdot \int \frac{2 u + 5}{4 u ^{2} - 1} d u

拡張

\frac{2 u + 5}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} + \frac{5}{4 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

= 2 (\int \frac{2 u}{4 u ^{2} - 1} d u + \int \frac{5}{4 u ^{2} - 1} d u)

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

\int \frac{5}{4 u ^{2} - 1} d u = - \frac{5}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

= 2 (\frac{1}{4} ln 4 u^{2} - 1 - \frac{5}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣))

代用を戻す

u = x - \frac{5}{2}

= 2 (\frac{1}{4} ln 4 (x - \frac{5}{2})^{2} - 1 - \frac{5}{4} (ln 2 (x - \frac{5}{2}) + 1 - ln 2 (x - \frac{5}{2}) - 1))

簡素化

2 (\frac{1}{4} ln 4 (x - \frac{5}{2})^{2} - 1 - \frac{5}{4} (ln 2 (x - \frac{5}{2}) + 1 - ln 2 (x - \frac{5}{2}) - 1)) : 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{5}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣))

= 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{5}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣))

定数を解答に追加する

= 2 (\frac{1}{4} ln 4 x^{2} - 20 x + 24 - \frac{5}{4} (ln ∣ 2 x - 4 ∣ - ln ∣ 2 x - 6 ∣)) + C

グラフ

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