integral of 1/((x+3)(x^2+x+1))

解

\int \frac{1}{( x + 3 ) ( x ^{2} + x + 1 )} d x

\frac{1}{7} ln ∣ x + 3 ∣ - \frac{1}{14} ln 4 x^{2} + 4 x + 4 + \frac{5}{7 3} arctan (\frac{1}{3} (2 x + 1)) + C

解答ステップ

\int \frac{1}{( x + 3 ) ( x ^{2} + x + 1 )} d x

以下の部分分数を得る:

\frac{1}{( x + 3 ) ( x ^{2} + x + 1 )} : \frac{1}{7 ( x + 3 )} + \frac{- x + 2}{7 ( x ^{2} + x + 1 )}

= \int \frac{1}{7 ( x + 3 )} + \frac{- x + 2}{7 ( 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}{7 ( x + 3 )} d x + \int \frac{- x + 2}{7 ( x ^{2} + x + 1 )} d x

\int \frac{1}{7 ( x + 3 )} d x = \frac{1}{7} ln ∣ x + 3 ∣

\int \frac{- x + 2}{7 ( x ^{2} + x + 1 )} d x = \frac{1}{14 3} (- 3 ln 4 x^{2} + 4 x + 4 + 10 arctan (\frac{2 x + 1}{3}))

= \frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{14 3} (- 3 ln 4 x^{2} + 4 x + 4 + 10 arctan (\frac{2 x + 1}{3}))

簡素化

\frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{14 3} (- 3 ln 4 x^{2} + 4 x + 4 + 10 arctan (\frac{2 x + 1}{3})) : \frac{1}{7} ln ∣ x + 3 ∣ - \frac{1}{14} ln 4 x^{2} + 4 x + 4 + \frac{5}{7 3} arctan (\frac{1}{3} (2 x + 1))

= \frac{1}{7} ln ∣ x + 3 ∣ - \frac{1}{14} ln 4 x^{2} + 4 x + 4 + \frac{5}{7 3} arctan (\frac{1}{3} (2 x + 1))

定数を解答に追加する

= \frac{1}{7} ln ∣ x + 3 ∣ - \frac{1}{14} ln 4 x^{2} + 4 x + 4 + \frac{5}{7 3} arctan (\frac{1}{3} (2 x + 1)) + C