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

解

\int \frac{x ^{2} + 2}{x ( x ^{2} + 5 x + 8 )} d x

\frac{1}{4} ln ∣ x ∣ + \frac{3}{8} ln 4 x^{2} + 20 x + 32 - \frac{25}{4 7} arctan (\frac{1}{7} (2 x + 5)) + C

解答ステップ

\int \frac{x ^{2} + 2}{x ( x ^{2} + 5 x + 8 )} d x

以下の部分分数を得る:

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

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

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

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{3 x - 5}{4 ( x ^{2} + 5 x + 8 )} d x = \frac{1}{8 7} (37 ln 4 x^{2} + 20 x + 32 - 50 arctan (\frac{2 x + 5}{7}))

= \frac{1}{4} ln ∣ x ∣ + \frac{1}{8 7} (37 ln 4 x^{2} + 20 x + 32 - 50 arctan (\frac{2 x + 5}{7}))

簡素化

\frac{1}{4} ln ∣ x ∣ + \frac{1}{8 7} (37 ln 4 x^{2} + 20 x + 32 - 50 arctan (\frac{2 x + 5}{7})) : \frac{1}{4} ln ∣ x ∣ + \frac{3}{8} ln 4 x^{2} + 20 x + 32 - \frac{25}{4 7} arctan (\frac{1}{7} (2 x + 5))

= \frac{1}{4} ln ∣ x ∣ + \frac{3}{8} ln 4 x^{2} + 20 x + 32 - \frac{25}{4 7} arctan (\frac{1}{7} (2 x + 5))

定数を解答に追加する

= \frac{1}{4} ln ∣ x ∣ + \frac{3}{8} ln 4 x^{2} + 20 x + 32 - \frac{25}{4 7} arctan (\frac{1}{7} (2 x + 5)) + C