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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

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

定数を解答に追加する

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