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

解

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

\frac{7}{9} ln ∣ x ∣ + \frac{2}{9} ln ∣ x + 3 ∣ - \frac{2}{3 x} + C

解答ステップ

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

拡張

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

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

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

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

\int \frac{2}{x ^{2} ( x + 3 )} d x = 2 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣)

= ln ∣ x + 3 ∣ + ln ∣ x ∣ - ln ∣ x + 3 ∣ + 2 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣)

簡素化

ln ∣ x + 3 ∣ + ln ∣ x ∣ - ln ∣ x + 3 ∣ + 2 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{3 x} + \frac{1}{9} ln ∣ x + 3 ∣) : \frac{7}{9} ln ∣ x ∣ + \frac{2}{9} ln ∣ x + 3 ∣ - \frac{2}{3 x}

= \frac{7}{9} ln ∣ x ∣ + \frac{2}{9} ln ∣ x + 3 ∣ - \frac{2}{3 x}

定数を解答に追加する

= \frac{7}{9} ln ∣ x ∣ + \frac{2}{9} ln ∣ x + 3 ∣ - \frac{2}{3 x} + C