integral of (4x^3-x)/(6x^2-2x-1)

解

\int \frac{4 x ^{3} - x}{6 x ^{2} - 2 x - 1} d x

\frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108} ln 36 x^{2} - 12 x - 6 - \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) + 1 + \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) - 1 + C

解答ステップ

\int \frac{4 x ^{3} - x}{6 x ^{2} - 2 x - 1} d x

長除法

\frac{4 x ^{3} - x}{6 x ^{2} - 2 x - 1} : \frac{2 x}{3} + \frac{2}{9} + \frac{x + 2}{9 ( 6 x ^{2} - 2 x - 1 )}

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

\int \frac{2 x}{3} d x = \frac{x ^{2}}{3}

\int \frac{2}{9} d x = \frac{2}{9} x

\int \frac{x + 2}{9 ( 6 x ^{2} - 2 x - 1 )} d x = \frac{1}{108 7} (7 ln 36 x^{2} - 12 x - 6 - 13 (ln \frac{6 x - 1}{7} + 1 - ln \frac{6 x - 1}{7} - 1))

= \frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108 7} (7 ln 36 x^{2} - 12 x - 6 - 13 (ln \frac{6 x - 1}{7} + 1 - ln \frac{6 x - 1}{7} - 1))

簡素化

\frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108 7} (7 ln 36 x^{2} - 12 x - 6 - 13 (ln \frac{6 x - 1}{7} + 1 - ln \frac{6 x - 1}{7} - 1)) : \frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108} ln 36 x^{2} - 12 x - 6 - \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) + 1 + \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) - 1

= \frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108} ln 36 x^{2} - 12 x - 6 - \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) + 1 + \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) - 1

定数を解答に追加する

= \frac{x ^{2}}{3} + \frac{2}{9} x + \frac{1}{108} ln 36 x^{2} - 12 x - 6 - \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) + 1 + \frac{13}{108 7} ln \frac{1}{7} (6 x - 1) - 1 + C