integral of (4-3x)/(5x^2+6x+18)

解

\int \frac{4 - 3 x}{5 x ^{2} + 6 x + 18} d x

\frac{29}{45} arctan (\frac{5}{9} x + \frac{1}{3}) - \frac{3}{10} ln \frac{25}{81} x^{2} + \frac{10 x}{27} + \frac{10}{9} + C

解答ステップ

\int \frac{4 - 3 x}{5 x ^{2} + 6 x + 18} d x

拡張

\frac{4 - 3 x}{5 x ^{2} + 6 x + 18} : \frac{4}{5 x ^{2} + 6 x + 18} - \frac{3 x}{5 x ^{2} + 6 x + 18}

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{4}{5 x ^{2} + 6 x + 18} d x - \int \frac{3 x}{5 x ^{2} + 6 x + 18} d x

\int \frac{4}{5 x ^{2} + 6 x + 18} d x = \frac{4}{9} arctan (\frac{5}{9} x + \frac{1}{3})

\int \frac{3 x}{5 x ^{2} + 6 x + 18} d x = \frac{1}{10} (3 ln \frac{10}{9} + \frac{25}{81} x^{2} + \frac{10 x}{27} - 2 arctan (\frac{5}{9} x + \frac{1}{3}))

= \frac{4}{9} arctan (\frac{5}{9} x + \frac{1}{3}) - \frac{1}{10} (3 ln \frac{10}{9} + \frac{25}{81} x^{2} + \frac{10 x}{27} - 2 arctan (\frac{5}{9} x + \frac{1}{3}))

簡素化

\frac{4}{9} arctan (\frac{5}{9} x + \frac{1}{3}) - \frac{1}{10} (3 ln \frac{10}{9} + \frac{25}{81} x^{2} + \frac{10 x}{27} - 2 arctan (\frac{5}{9} x + \frac{1}{3})) : \frac{29}{45} arctan (\frac{5}{9} x + \frac{1}{3}) - \frac{3}{10} ln \frac{25}{81} x^{2} + \frac{10 x}{27} + \frac{10}{9}

= \frac{29}{45} arctan (\frac{5}{9} x + \frac{1}{3}) - \frac{3}{10} ln \frac{25}{81} x^{2} + \frac{10 x}{27} + \frac{10}{9}

定数を解答に追加する

= \frac{29}{45} arctan (\frac{5}{9} x + \frac{1}{3}) - \frac{3}{10} ln \frac{25}{81} x^{2} + \frac{10 x}{27} + \frac{10}{9} + C