integral of ((3x^3-5x^2+10x-3))/(3x+1)

解

\int \frac{( 3 x ^{3} - 5 x ^{2} + 10 x - 3 )}{3 x + 1} d x

- \frac{5}{6} x^{2} + \frac{35}{9} x - \frac{62}{27} ln ∣ 3 x + 1 ∣ + \frac{1}{162} (2 (3 x + 1)^{3} - 9 (3 x + 1)^{2} + 18 (3 x + 1) - 6 ln ∣ 3 x + 1 ∣) + \frac{5}{18} + \frac{10}{9} + C

解答ステップ

\int \frac{3 x ^{3} - 5 x ^{2} + 10 x - 3}{3 x + 1} d x

拡張

\frac{3 x ^{3} - 5 x ^{2} + 10 x - 3}{3 x + 1} : \frac{3 x ^{3}}{3 x + 1} - \frac{5 x ^{2}}{3 x + 1} + \frac{10 x}{3 x + 1} - \frac{3}{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{3 x ^{3}}{3 x + 1} d x - \int \frac{5 x ^{2}}{3 x + 1} d x + \int \frac{10 x}{3 x + 1} d x - \int \frac{3}{3 x + 1} d x

\int \frac{3 x ^{3}}{3 x + 1} d x = \frac{1}{162} (2 (3 x + 1)^{3} - 9 (3 x + 1)^{2} + 18 (3 x + 1) - 6 ln ∣ 3 x + 1 ∣)

\int \frac{5 x ^{2}}{3 x + 1} d x = \frac{5}{54} (9 x^{2} - 6 x + 2 ln ∣ 3 x + 1 ∣ - 3)

\int \frac{10 x}{3 x + 1} d x = \frac{10}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣)

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

= \frac{1}{162} (2 (3 x + 1)^{3} - 9 (3 x + 1)^{2} + 18 (3 x + 1) - 6 ln ∣ 3 x + 1 ∣) - \frac{5}{54} (9 x^{2} - 6 x + 2 ln ∣ 3 x + 1 ∣ - 3) + \frac{10}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣) - ln ∣ 3 x + 1 ∣

簡素化

\frac{1}{162} (2 (3 x + 1)^{3} - 9 (3 x + 1)^{2} + 18 (3 x + 1) - 6 ln ∣ 3 x + 1 ∣) - \frac{5}{54} (9 x^{2} - 6 x + 2 ln ∣ 3 x + 1 ∣ - 3) + \frac{10}{9} (3 x + 1 - ln ∣ 3 x + 1 ∣) - ln ∣ 3 x + 1 ∣ : - \frac{5}{6} x^{2} + \frac{35}{9} x - \frac{62}{27} ln ∣ 3 x + 1 ∣ + \frac{1}{162} (2 (3 x + 1)^{3} - 9 (3 x + 1)^{2} + 18 (3 x + 1) - 6 ln ∣ 3 x + 1 ∣) + \frac{5}{18} + \frac{10}{9}

= - \frac{5}{6} x^{2} + \frac{35}{9} x - \frac{62}{27} ln ∣ 3 x + 1 ∣ + \frac{1}{162} (2 (3 x + 1)^{3} - 9 (3 x + 1)^{2} + 18 (3 x + 1) - 6 ln ∣ 3 x + 1 ∣) + \frac{5}{18} + \frac{10}{9}

定数を解答に追加する

= - \frac{5}{6} x^{2} + \frac{35}{9} x - \frac{62}{27} ln ∣ 3 x + 1 ∣ + \frac{1}{162} (2 (3 x + 1)^{3} - 9 (3 x + 1)^{2} + 18 (3 x + 1) - 6 ln ∣ 3 x + 1 ∣) + \frac{5}{18} + \frac{10}{9} + C