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

解

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

\frac{x ^{3} + x ^{2} + x - 2}{x ( x - 1 )} + ln ∣ x - 1 ∣ - \frac{1}{x - 1} + ln ∣ x ∣ - \frac{4}{x} + C

解答ステップ

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

部分積分を適用する

= \frac{x ^{3} + x ^{2} + x - 2}{x ( x - 1 )} - \int - \frac{2 x ^{3} + 2 x ^{2} - 7 x + 4}{x ^{2} ( x - 1 ) ^{2}} d x

\int - \frac{2 x ^{3} + 2 x ^{2} - 7 x + 4}{x ^{2} ( x - 1 ) ^{2}} d x = - ln ∣ x - 1 ∣ + \frac{1}{x - 1} - ln ∣ x ∣ + \frac{4}{x}

= \frac{x ^{3} + x ^{2} + x - 2}{x ( x - 1 )} - (- ln ∣ x - 1 ∣ + \frac{1}{x - 1} - ln ∣ x ∣ + \frac{4}{x})

簡素化

= \frac{x ^{3} + x ^{2} + x - 2}{x ( x - 1 )} + ln ∣ x - 1 ∣ - \frac{1}{x - 1} + ln ∣ x ∣ - \frac{4}{x}

定数を解答に追加する

= \frac{x ^{3} + x ^{2} + x - 2}{x ( x - 1 )} + ln ∣ x - 1 ∣ - \frac{1}{x - 1} + ln ∣ x ∣ - \frac{4}{x} + C