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

解

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

- \frac{x}{100 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{9}{125} arctan (x) - \frac{13}{125} ln ∣ x + 2 ∣ - \frac{73}{1000} ln x^{2} + 1 + \frac{7}{100 ( x ^{2} + 1 )} + \frac{1}{25 ( x + 2 )} + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( x + 1 ) ( x ^{2} + 1 ) ^{2} ( x + 2 ) ^{2}} : \frac{1}{4 ( x + 1 )} + \frac{- 73 x + 41}{500 ( x ^{2} + 1 )} + \frac{- 7 x - 1}{50 ( x ^{2} + 1 ) ^{2}} - \frac{13}{125 ( x + 2 )} - \frac{1}{25 ( x + 2 ) ^{2}}

= \int \frac{1}{4 ( x + 1 )} + \frac{- 73 x + 41}{500 ( x ^{2} + 1 )} + \frac{- 7 x - 1}{50 ( x ^{2} + 1 ) ^{2}} - \frac{13}{125 ( x + 2 )} - \frac{1}{25 ( x + 2 ) ^{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{- 73 x + 41}{500 ( x ^{2} + 1 )} d x + \int \frac{- 7 x - 1}{50 ( x ^{2} + 1 ) ^{2}} d x - \int \frac{13}{125 ( x + 2 )} d x - \int \frac{1}{25 ( x + 2 ) ^{2}} d x

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

\int \frac{- 73 x + 41}{500 ( x ^{2} + 1 )} d x = \frac{1}{500} (- \frac{73}{2} ln x^{2} + 1 + 41 arctan (x))

\int \frac{- 7 x - 1}{50 ( x ^{2} + 1 ) ^{2}} d x = \frac{1}{50} (\frac{7}{2 ( x ^{2} + 1 )} - \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}))

\int \frac{13}{125 ( x + 2 )} d x = \frac{13}{125} ln ∣ x + 2 ∣

\int \frac{1}{25 ( x + 2 ) ^{2}} d x = - \frac{1}{25 ( x + 2 )}

= \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{500} (- \frac{73}{2} ln x^{2} + 1 + 41 arctan (x)) + \frac{1}{50} (\frac{7}{2 ( x ^{2} + 1 )} - \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}})) - \frac{13}{125} ln ∣ x + 2 ∣ - (- \frac{1}{25 ( x + 2 )})

簡素化

\frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{500} (- \frac{73}{2} ln x^{2} + 1 + 41 arctan (x)) + \frac{1}{50} (\frac{7}{2 ( x ^{2} + 1 )} - \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}})) - \frac{13}{125} ln ∣ x + 2 ∣ - (- \frac{1}{25 ( x + 2 )}) : - \frac{x}{100 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{9}{125} arctan (x) - \frac{13}{125} ln ∣ x + 2 ∣ - \frac{73}{1000} ln x^{2} + 1 + \frac{7}{100 ( x ^{2} + 1 )} + \frac{1}{25 ( x + 2 )}

= - \frac{x}{100 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{9}{125} arctan (x) - \frac{13}{125} ln ∣ x + 2 ∣ - \frac{73}{1000} ln x^{2} + 1 + \frac{7}{100 ( x ^{2} + 1 )} + \frac{1}{25 ( x + 2 )}

定数を解答に追加する

= - \frac{x}{100 ( x ^{2} + 1 )} + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{9}{125} arctan (x) - \frac{13}{125} ln ∣ x + 2 ∣ - \frac{73}{1000} ln x^{2} + 1 + \frac{7}{100 ( x ^{2} + 1 )} + \frac{1}{25 ( x + 2 )} + C