integral of (6x^3+4x^2-45x-33)/(x^2-9)

解

\int \frac{6 x ^{3} + 4 x ^{2} - 45 x - 33}{x ^{2} - 9} d x

3 x^{2} + 4 x + \frac{9}{2} ln x^{2} - 9 - \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1 - 27 + C

解答ステップ

\int \frac{6 x ^{3} + 4 x ^{2} - 45 x - 33}{x ^{2} - 9} d x

拡張

\frac{6 x ^{3} + 4 x ^{2} - 45 x - 33}{x ^{2} - 9} : \frac{6 x ^{3}}{x ^{2} - 9} + \frac{4 x ^{2}}{x ^{2} - 9} - \frac{45 x}{x ^{2} - 9} - \frac{33}{x ^{2} - 9}

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{6 x ^{3}}{x ^{2} - 9} d x + \int \frac{4 x ^{2}}{x ^{2} - 9} d x - \int \frac{45 x}{x ^{2} - 9} d x - \int \frac{33}{x ^{2} - 9} d x

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

\int \frac{4 x ^{2}}{x ^{2} - 9} d x = - 12 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3})

\int \frac{45 x}{x ^{2} - 9} d x = \frac{45}{2} ln x^{2} - 9

\int \frac{33}{x ^{2} - 9} d x = - 11 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1)

= 3 (x^{2} - 9 + 9 ln x^{2} - 9) - 12 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3}) - \frac{45}{2} ln x^{2} - 9 - (- 11 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1))

簡素化

3 (x^{2} - 9 + 9 ln x^{2} - 9) - 12 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1 - \frac{x}{3}) - \frac{45}{2} ln x^{2} - 9 - (- 11 (\frac{1}{2} ln \frac{x}{3} + 1 - \frac{1}{2} ln \frac{x}{3} - 1)) : 3 x^{2} + 4 x + \frac{9}{2} ln x^{2} - 9 - \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1 - 27

= 3 x^{2} + 4 x + \frac{9}{2} ln x^{2} - 9 - \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1 - 27

定数を解答に追加する

= 3 x^{2} + 4 x + \frac{9}{2} ln x^{2} - 9 - \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1 - 27 + C