integral of (3x^2-x+4)/(x^3+2x^2+6x)

解

\int \frac{3 x ^{2} - x + 4}{x ^{3} + 2 x ^{2} + 6 x} d x

\frac{7}{6} ln x^{2} + 2 x + 6 + \frac{2}{3} ln ∣ x ∣ - \frac{11}{3 5} arctan (\frac{1}{5} (x + 1)) - \frac{1}{5} arctan (\frac{1}{5} (x + 1)) + C

解答ステップ

\int \frac{3 x ^{2} - x + 4}{x ^{3} + 2 x ^{2} + 6 x} d x

拡張

\frac{3 x ^{2} - x + 4}{x ^{3} + 2 x ^{2} + 6 x} : \frac{3 x ^{2}}{x ^{3} + 2 x ^{2} + 6 x} - \frac{x}{x ^{3} + 2 x ^{2} + 6 x} + \frac{4}{x ^{3} + 2 x ^{2} + 6 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{3 x ^{2}}{x ^{3} + 2 x ^{2} + 6 x} d x - \int \frac{x}{x ^{3} + 2 x ^{2} + 6 x} d x + \int \frac{4}{x ^{3} + 2 x ^{2} + 6 x} d x

\int \frac{3 x ^{2}}{x ^{3} + 2 x ^{2} + 6 x} d x = 3 (\frac{1}{2} ln x^{2} + 2 x + 6 - \frac{1}{5} arctan (\frac{1}{5} (x + 1)))

\int \frac{x}{x ^{3} + 2 x ^{2} + 6 x} d x = \frac{1}{5} arctan (\frac{x + 1}{5})

\int \frac{4}{x ^{3} + 2 x ^{2} + 6 x} d x = 4 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 2 x + 6 - \frac{1}{6 5} arctan (\frac{1}{5} (x + 1)))

= 3 (\frac{1}{2} ln x^{2} + 2 x + 6 - \frac{1}{5} arctan (\frac{1}{5} (x + 1))) - \frac{1}{5} arctan (\frac{x + 1}{5}) + 4 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 2 x + 6 - \frac{1}{6 5} arctan (\frac{1}{5} (x + 1)))

簡素化

3 (\frac{1}{2} ln x^{2} + 2 x + 6 - \frac{1}{5} arctan (\frac{1}{5} (x + 1))) - \frac{1}{5} arctan (\frac{x + 1}{5}) + 4 (\frac{1}{6} ln ∣ x ∣ - \frac{1}{12} ln x^{2} + 2 x + 6 - \frac{1}{6 5} arctan (\frac{1}{5} (x + 1))) : \frac{7}{6} ln x^{2} + 2 x + 6 + \frac{2}{3} ln ∣ x ∣ - \frac{11}{3 5} arctan (\frac{1}{5} (x + 1)) - \frac{1}{5} arctan (\frac{1}{5} (x + 1))

= \frac{7}{6} ln x^{2} + 2 x + 6 + \frac{2}{3} ln ∣ x ∣ - \frac{11}{3 5} arctan (\frac{1}{5} (x + 1)) - \frac{1}{5} arctan (\frac{1}{5} (x + 1))

定数を解答に追加する

= \frac{7}{6} ln x^{2} + 2 x + 6 + \frac{2}{3} ln ∣ x ∣ - \frac{11}{3 5} arctan (\frac{1}{5} (x + 1)) - \frac{1}{5} arctan (\frac{1}{5} (x + 1)) + C