integral of (3x^2+8x-12)/(x^3+7x+12x)

解

\int \frac{3 x ^{2} + 8 x - 12}{x ^{3} + 7 x + 12 x} d x

\frac{69}{38} ln x^{2} + 19 - \frac{12}{19} ln ∣ x ∣ + \frac{8}{19} arctan (\frac{x}{19}) + C

解答ステップ

\int \frac{3 x ^{2} + 8 x - 12}{x ^{3} + 7 x + 12 x} d x

拡張

\frac{3 x ^{2} + 8 x - 12}{x ^{3} + 7 x + 12 x} : \frac{3 x ^{2}}{x ^{3} + 7 x + 12 x} + \frac{8 x}{x ^{3} + 7 x + 12 x} - \frac{12}{x ^{3} + 7 x + 12 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} + 7 x + 12 x} d x + \int \frac{8 x}{x ^{3} + 7 x + 12 x} d x - \int \frac{12}{x ^{3} + 7 x + 12 x} d x

\int \frac{3 x ^{2}}{x ^{3} + 7 x + 12 x} d x = \frac{3}{2} ln x^{2} + 19

\int \frac{8 x}{x ^{3} + 7 x + 12 x} d x = \frac{8}{19} arctan (\frac{x}{19})

\int \frac{12}{x ^{3} + 7 x + 12 x} d x = 12 (\frac{1}{19} ln ∣ x ∣ - \frac{1}{38} ln x^{2} + 19)

= \frac{3}{2} ln x^{2} + 19 + \frac{8}{19} arctan (\frac{x}{19}) - 12 (\frac{1}{19} ln ∣ x ∣ - \frac{1}{38} ln x^{2} + 19)

簡素化

\frac{3}{2} ln x^{2} + 19 + \frac{8}{19} arctan (\frac{x}{19}) - 12 (\frac{1}{19} ln ∣ x ∣ - \frac{1}{38} ln x^{2} + 19) : \frac{69}{38} ln x^{2} + 19 - \frac{12}{19} ln ∣ x ∣ + \frac{8}{19} arctan (\frac{x}{19})

= \frac{69}{38} ln x^{2} + 19 - \frac{12}{19} ln ∣ x ∣ + \frac{8}{19} arctan (\frac{x}{19})

定数を解答に追加する

= \frac{69}{38} ln x^{2} + 19 - \frac{12}{19} ln ∣ x ∣ + \frac{8}{19} arctan (\frac{x}{19}) + C