integral of (3x^2+2x-10)/(x^3+x^2-10x+8)

解

\int \frac{3 x ^{2} + 2 x - 10}{x ^{3} + x ^{2} - 10 x + 8} d x

ln ∣ x - 1 ∣ + ln ∣ x + 4 ∣ + ln ∣ x - 2 ∣ + C

解答ステップ

\int \frac{3 x ^{2} + 2 x - 10}{x ^{3} + x ^{2} - 10 x + 8} d x

拡張

\frac{3 x ^{2} + 2 x - 10}{x ^{3} + x ^{2} - 10 x + 8} : \frac{3 x ^{2}}{x ^{3} + x ^{2} - 10 x + 8} + \frac{2 x}{x ^{3} + x ^{2} - 10 x + 8} - \frac{10}{x ^{3} + x ^{2} - 10 x + 8}

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} + x ^{2} - 10 x + 8} d x + \int \frac{2 x}{x ^{3} + x ^{2} - 10 x + 8} d x - \int \frac{10}{x ^{3} + x ^{2} - 10 x + 8} d x

\int \frac{3 x ^{2}}{x ^{3} + x ^{2} - 10 x + 8} d x = - \frac{3}{5} ln ∣ x - 1 ∣ + 2 ln ∣ x - 2 ∣ + \frac{8}{5} ln ∣ x + 4 ∣

\int \frac{2 x}{x ^{3} + x ^{2} - 10 x + 8} d x = 2 (- \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣ - \frac{2}{15} ln ∣ x + 4 ∣)

\int \frac{10}{x ^{3} + x ^{2} - 10 x + 8} d x = 10 (- \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x - 2 ∣ + \frac{1}{30} ln ∣ x + 4 ∣)

= - \frac{3}{5} ln ∣ x - 1 ∣ + 2 ln ∣ x - 2 ∣ + \frac{8}{5} ln ∣ x + 4 ∣ + 2 (- \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣ - \frac{2}{15} ln ∣ x + 4 ∣) - 10 (- \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x - 2 ∣ + \frac{1}{30} ln ∣ x + 4 ∣)

簡素化

- \frac{3}{5} ln ∣ x - 1 ∣ + 2 ln ∣ x - 2 ∣ + \frac{8}{5} ln ∣ x + 4 ∣ + 2 (- \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{3} ln ∣ x - 2 ∣ - \frac{2}{15} ln ∣ x + 4 ∣) - 10 (- \frac{1}{5} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x - 2 ∣ + \frac{1}{30} ln ∣ x + 4 ∣) : ln ∣ x - 1 ∣ + ln ∣ x + 4 ∣ + ln ∣ x - 2 ∣

= ln ∣ x - 1 ∣ + ln ∣ x + 4 ∣ + ln ∣ x - 2 ∣

定数を解答に追加する

= ln ∣ x - 1 ∣ + ln ∣ x + 4 ∣ + ln ∣ x - 2 ∣ + C