integral of (x^2+2)/((x+1)(x^2+4))

解

\int \frac{x ^{2} + 2}{( x + 1 ) ( x ^{2} + 4 )} d x

\frac{3}{5} ln ∣ x + 1 ∣ + \frac{1}{5} (ln x^{2} + 4 - arctan (\frac{x}{2})) + C

解答ステップ

\int \frac{x ^{2} + 2}{( x + 1 ) ( x ^{2} + 4 )} d x

以下の部分分数を得る:

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

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

\int \frac{3}{5 ( x + 1 )} d x = \frac{3}{5} ln ∣ x + 1 ∣

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

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

定数を解答に追加する

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