integral of (3x^2-21x+32)/(x^3-8x^2+16x)

解

\int \frac{3 x ^{2} - 21 x + 32}{x ^{3} - 8 x ^{2} + 16 x} d x

ln ∣ x - 4 ∣ + \frac{1}{x - 4} + 2 ln ∣ x ∣ + C

解答ステップ

\int \frac{3 x ^{2} - 21 x + 32}{x ^{3} - 8 x ^{2} + 16 x} d x

拡張

\frac{3 x ^{2} - 21 x + 32}{x ^{3} - 8 x ^{2} + 16 x} : \frac{3 x ^{2}}{x ^{3} - 8 x ^{2} + 16 x} - \frac{21 x}{x ^{3} - 8 x ^{2} + 16 x} + \frac{32}{x ^{3} - 8 x ^{2} + 16 x}

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{3 x ^{2}}{x ^{3} - 8 x ^{2} + 16 x} d x - \int \frac{21 x}{x ^{3} - 8 x ^{2} + 16 x} d x + \int \frac{32}{x ^{3} - 8 x ^{2} + 16 x} d x

\int \frac{3 x ^{2}}{x ^{3} - 8 x ^{2} + 16 x} d x = 3 (ln ∣ x - 4 ∣ - \frac{4}{x - 4})

\int \frac{21 x}{x ^{3} - 8 x ^{2} + 16 x} d x = - \frac{21}{x - 4}

\int \frac{32}{x ^{3} - 8 x ^{2} + 16 x} d x = 32 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )})

= 3 (ln ∣ x - 4 ∣ - \frac{4}{x - 4}) - (- \frac{21}{x - 4}) + 32 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )})

簡素化

3 (ln ∣ x - 4 ∣ - \frac{4}{x - 4}) - (- \frac{21}{x - 4}) + 32 (\frac{1}{16} ln ∣ x ∣ - \frac{1}{16} ln ∣ x - 4 ∣ - \frac{1}{4 ( x - 4 )}) : ln ∣ x - 4 ∣ + \frac{1}{x - 4} + 2 ln ∣ x ∣

= ln ∣ x - 4 ∣ + \frac{1}{x - 4} + 2 ln ∣ x ∣

定数を解答に追加する

= ln ∣ x - 4 ∣ + \frac{1}{x - 4} + 2 ln ∣ x ∣ + C