integral of (2x+3)/(9x^2-12x+8)

解

\int \frac{2 x + 3}{9 x ^{2} - 12 x + 8} d x

\frac{1}{9} (ln 9 x^{2} - 12 x + 8 + 2 arctan (\frac{3 x - 2}{2})) + \frac{1}{2} arctan (\frac{3}{2} x - 1) + C

解答ステップ

\int \frac{2 x + 3}{9 x ^{2} - 12 x + 8} d x

拡張

\frac{2 x + 3}{9 x ^{2} - 12 x + 8} : \frac{2 x}{9 x ^{2} - 12 x + 8} + \frac{3}{9 x ^{2} - 12 x + 8}

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}{9 x ^{2} - 12 x + 8} d x + \int \frac{3}{9 x ^{2} - 12 x + 8} d x

\int \frac{2 x}{9 x ^{2} - 12 x + 8} d x = \frac{1}{9} (ln 9 x^{2} - 12 x + 8 + 2 arctan (\frac{3 x - 2}{2}))

\int \frac{3}{9 x ^{2} - 12 x + 8} d x = \frac{1}{2} arctan (\frac{3}{2} x - 1)

= \frac{1}{9} (ln 9 x^{2} - 12 x + 8 + 2 arctan (\frac{3 x - 2}{2})) + \frac{1}{2} arctan (\frac{3}{2} x - 1)

定数を解答に追加する

= \frac{1}{9} (ln 9 x^{2} - 12 x + 8 + 2 arctan (\frac{3 x - 2}{2})) + \frac{1}{2} arctan (\frac{3}{2} x - 1) + C