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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

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

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

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

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

簡素化

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

定数を解答に追加する

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