integral of (5x-2)/(x^2+6x+8)

解

\int \frac{5 x - 2}{x ^{2} + 6 x + 8} d x

\frac{5}{2} ln x^{2} + 6 x + 8 + \frac{17}{2} ln ∣ x + 4 ∣ - \frac{17}{2} ln ∣ x + 2 ∣ + C

解答ステップ

\int \frac{5 x - 2}{x ^{2} + 6 x + 8} d x

拡張

\frac{5 x - 2}{x ^{2} + 6 x + 8} : \frac{5 x}{x ^{2} + 6 x + 8} - \frac{2}{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{5 x}{x ^{2} + 6 x + 8} d x - \int \frac{2}{x ^{2} + 6 x + 8} d x

\int \frac{5 x}{x ^{2} + 6 x + 8} d x = 5 (\frac{1}{2} ln x^{2} + 6 x + 8 + 3 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 2 ∣))

\int \frac{2}{x ^{2} + 6 x + 8} d x = ln ∣ x + 2 ∣ - ln ∣ x + 4 ∣

= 5 (\frac{1}{2} ln x^{2} + 6 x + 8 + 3 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 2 ∣)) - (ln ∣ x + 2 ∣ - ln ∣ x + 4 ∣)

簡素化

5 (\frac{1}{2} ln x^{2} + 6 x + 8 + 3 (\frac{1}{2} ln ∣ x + 4 ∣ - \frac{1}{2} ln ∣ x + 2 ∣)) - (ln ∣ x + 2 ∣ - ln ∣ x + 4 ∣) : \frac{5}{2} ln x^{2} + 6 x + 8 + \frac{17}{2} ln ∣ x + 4 ∣ - \frac{17}{2} ln ∣ x + 2 ∣

= \frac{5}{2} ln x^{2} + 6 x + 8 + \frac{17}{2} ln ∣ x + 4 ∣ - \frac{17}{2} ln ∣ x + 2 ∣

定数を解答に追加する

= \frac{5}{2} ln x^{2} + 6 x + 8 + \frac{17}{2} ln ∣ x + 4 ∣ - \frac{17}{2} ln ∣ x + 2 ∣ + C