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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{2}{5 ( x + 2 )} d x = \frac{2}{5} ln ∣ x + 2 ∣

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

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

定数を解答に追加する

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