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

解

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

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

解答ステップ

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

拡張

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

\int \frac{x ^{2}}{x ^{2} + x} d x = \frac{x ^{2}}{x + 1} - ln ∣ x + 1 ∣ - \frac{1}{x + 1}

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

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

簡素化

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

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

定数を解答に追加する

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