integral of (x^2+x+1)/(x^3+2x^2)

解

\int \frac{x ^{2} + x + 1}{x ^{3} + 2 x ^{2}} d x

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

解答ステップ

\int \frac{x ^{2} + x + 1}{x ^{3} + 2 x ^{2}} d x

拡張

\frac{x ^{2} + x + 1}{x ^{3} + 2 x ^{2}} : \frac{x ^{2}}{x ^{3} + 2 x ^{2}} + \frac{x}{x ^{3} + 2 x ^{2}} + \frac{1}{x ^{3} + 2 x ^{2}}

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 ^{3} + 2 x ^{2}} d x + \int \frac{x}{x ^{3} + 2 x ^{2}} d x + \int \frac{1}{x ^{3} + 2 x ^{2}} d x

\int \frac{x ^{2}}{x ^{3} + 2 x ^{2}} d x = \frac{x}{x + 2} + ln ∣ x + 2 ∣ + \frac{2}{x + 2}

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

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

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

簡素化

\frac{x}{x + 2} + ln ∣ x + 2 ∣ + \frac{2}{x + 2} + \frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 2 ∣ - \frac{1}{4} ln ∣ x ∣ - \frac{1}{2 x} + \frac{1}{4} ln ∣ x + 2 ∣ : \frac{1}{4} ln ∣ x ∣ + \frac{3}{4} ln ∣ x + 2 ∣ + \frac{x}{x + 2} + \frac{2}{x + 2} - \frac{1}{2 x}

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

定数を解答に追加する

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