integral of ((2x^2-5x+1))/((x+2)(x^2+8))

解

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

\frac{19}{12} ln ∣ x + 2 ∣ + \frac{5}{12 2} (\frac{1}{2} ln \frac{x ^{2}}{8} + 1 - 7 arctan (\frac{x}{2 2})) + C

解答ステップ

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

以下の部分分数を得る:

\frac{2 x ^{2} - 5 x + 1}{( x + 2 ) ( x ^{2} + 8 )} : \frac{19}{12 ( x + 2 )} + \frac{5 x - 70}{12 ( x ^{2} + 8 )}

= \int \frac{19}{12 ( x + 2 )} + \frac{5 x - 70}{12 ( x ^{2} + 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{19}{12 ( x + 2 )} d x + \int \frac{5 x - 70}{12 ( x ^{2} + 8 )} d x

\int \frac{19}{12 ( x + 2 )} d x = \frac{19}{12} ln ∣ x + 2 ∣

\int \frac{5 x - 70}{12 ( x ^{2} + 8 )} d x = \frac{5}{12 2} (\frac{1}{2} ln \frac{x ^{2}}{8} + 1 - 7 arctan (\frac{x}{2 2}))

= \frac{19}{12} ln ∣ x + 2 ∣ + \frac{5}{12 2} (\frac{1}{2} ln \frac{x ^{2}}{8} + 1 - 7 arctan (\frac{x}{2 2}))

定数を解答に追加する

= \frac{19}{12} ln ∣ x + 2 ∣ + \frac{5}{12 2} (\frac{1}{2} ln \frac{x ^{2}}{8} + 1 - 7 arctan (\frac{x}{2 2})) + C