integral of 1/(x^5+x^4+2x^3+2x^2+x+1)

解

\int \frac{1}{x ^{5} + x ^{4} + 2 x ^{3} + 2 x ^{2} + x + 1} d x

\frac{x}{4 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{4 ( x ^{2} + 1 )} + C

解答ステップ

\int \frac{1}{x ^{5} + x ^{4} + 2 x ^{3} + 2 x ^{2} + x + 1} d x

以下の部分分数を得る:

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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

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

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

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

簡素化

\frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln x^{2} + 1 + arctan (x)) + \frac{1}{2} (\frac{x}{2 ( x ^{2} + 1 )} + \frac{1}{2} arctan (x) + \frac{1}{2 ( x ^{2} + 1 )}) : \frac{x}{4 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{4 ( x ^{2} + 1 )}

= \frac{x}{4 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{4 ( x ^{2} + 1 )}

定数を解答に追加する

= \frac{x}{4 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{8} ln x^{2} + 1 + \frac{1}{2} arctan (x) + \frac{1}{4 ( x ^{2} + 1 )} + C