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

解

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

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

解答ステップ

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

拡張

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

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 ^{2} ( x + 1 )} d x + \int \frac{2 x}{x ^{2} ( x + 1 )} d x + \int \frac{1}{x ^{2} ( x + 1 )} d x

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

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

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

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

簡素化

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

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

定数を解答に追加する

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