integral of (x+3)/(3x^2-6x+8)

解

\int \frac{x + 3}{3 x ^{2} - 6 x + 8} d x

\frac{1}{6} ln 3 x^{2} - 6 x + 8 + \frac{1}{15} arctan (\frac{3}{5} (x - 1)) + \frac{3}{5} arctan (\frac{3}{5} (x - 1)) + C

解答ステップ

\int \frac{x + 3}{3 x ^{2} - 6 x + 8} d x

拡張

\frac{x + 3}{3 x ^{2} - 6 x + 8} : \frac{x}{3 x ^{2} - 6 x + 8} + \frac{3}{3 x ^{2} - 6 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{x}{3 x ^{2} - 6 x + 8} d x + \int \frac{3}{3 x ^{2} - 6 x + 8} d x

\int \frac{x}{3 x ^{2} - 6 x + 8} d x = \frac{1}{6} ln 3 x^{2} - 6 x + 8 + \frac{1}{15} arctan (\frac{3}{5} (x - 1))

\int \frac{3}{3 x ^{2} - 6 x + 8} d x = \frac{3}{5} arctan (\frac{3}{5} (x - 1))

= \frac{1}{6} ln 3 x^{2} - 6 x + 8 + \frac{1}{15} arctan (\frac{3}{5} (x - 1)) + \frac{3}{5} arctan (\frac{3}{5} (x - 1))

定数を解答に追加する

= \frac{1}{6} ln 3 x^{2} - 6 x + 8 + \frac{1}{15} arctan (\frac{3}{5} (x - 1)) + \frac{3}{5} arctan (\frac{3}{5} (x - 1)) + C