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

解

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

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

解答ステップ

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

簡素化

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

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

以下の部分分数を得る:

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

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

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

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

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

定数を解答に追加する

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