integral of (3x+2)/(x^2-4x+20)

解

\int \frac{3 x + 2}{x ^{2} - 4 x + 20} d x

\frac{3}{2} ln \frac{1}{16} (x - 2)^{2} + 1 + 2 arctan (\frac{1}{4} (x - 2)) + C

解答ステップ

\int \frac{3 x + 2}{x ^{2} - 4 x + 20} d x

拡張

\frac{3 x + 2}{x ^{2} - 4 x + 20} : \frac{3 x}{x ^{2} - 4 x + 20} + \frac{2}{x ^{2} - 4 x + 20}

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{3 x}{x ^{2} - 4 x + 20} d x + \int \frac{2}{x ^{2} - 4 x + 20} d x

\int \frac{3 x}{x ^{2} - 4 x + 20} d x = \frac{3}{2} (ln \frac{1}{16} (x - 2)^{2} + 1 + arctan (\frac{1}{4} (x - 2)))

\int \frac{2}{x ^{2} - 4 x + 20} d x = \frac{1}{2} arctan (\frac{1}{4} (x - 2))

= \frac{3}{2} (ln \frac{1}{16} (x - 2)^{2} + 1 + arctan (\frac{1}{4} (x - 2))) + \frac{1}{2} arctan (\frac{1}{4} (x - 2))

簡素化

\frac{3}{2} (ln \frac{1}{16} (x - 2)^{2} + 1 + arctan (\frac{1}{4} (x - 2))) + \frac{1}{2} arctan (\frac{1}{4} (x - 2)) : \frac{3}{2} ln \frac{1}{16} (x - 2)^{2} + 1 + 2 arctan (\frac{1}{4} (x - 2))

= \frac{3}{2} ln \frac{1}{16} (x - 2)^{2} + 1 + 2 arctan (\frac{1}{4} (x - 2))

定数を解答に追加する

= \frac{3}{2} ln \frac{1}{16} (x - 2)^{2} + 1 + 2 arctan (\frac{1}{4} (x - 2)) + C