integral of (x^3+4x^2+x-1)/(x^3+x^2)

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

\int 1 d x = x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

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

簡素化

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

定数を解答に追加する

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