integral of (12x^2-4x+7)/(3x+1)

解

\int \frac{12 x ^{2} - 4 x + 7}{3 x + 1} d x

2 x^{2} - \frac{8}{3} x + \frac{29}{9} ln ∣ 3 x + 1 ∣ - \frac{2}{3} - \frac{4}{9} + C

解答ステップ

\int \frac{12 x ^{2} - 4 x + 7}{3 x + 1} d x

拡張

\frac{12 x ^{2} - 4 x + 7}{3 x + 1} : \frac{12 x ^{2}}{3 x + 1} - \frac{4 x}{3 x + 1} + \frac{7}{3 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{12 x ^{2}}{3 x + 1} d x - \int \frac{4 x}{3 x + 1} d x + \int \frac{7}{3 x + 1} d x

\int \frac{12 x ^{2}}{3 x + 1} d x = \frac{2}{9} (9 x^{2} - 6 x + 2 ln ∣ 3 x + 1 ∣ - 3)

\int \frac{4 x}{3 x + 1} d x = \frac{4}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣)

\int \frac{7}{3 x + 1} d x = \frac{7}{3} ln ∣ 3 x + 1 ∣

= \frac{2}{9} (9 x^{2} - 6 x + 2 ln ∣ 3 x + 1 ∣ - 3) - \frac{4}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣) + \frac{7}{3} ln ∣ 3 x + 1 ∣

簡素化

\frac{2}{9} (9 x^{2} - 6 x + 2 ln ∣ 3 x + 1 ∣ - 3) - \frac{4}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣) + \frac{7}{3} ln ∣ 3 x + 1 ∣ : 2 x^{2} - \frac{8}{3} x + \frac{29}{9} ln ∣ 3 x + 1 ∣ - \frac{2}{3} - \frac{4}{9}

= 2 x^{2} - \frac{8}{3} x + \frac{29}{9} ln ∣ 3 x + 1 ∣ - \frac{2}{3} - \frac{4}{9}

定数を解答に追加する

= 2 x^{2} - \frac{8}{3} x + \frac{29}{9} ln ∣ 3 x + 1 ∣ - \frac{2}{3} - \frac{4}{9} + C