integral of 1/(x^8(1+x^2))

解

\int \frac{1}{x ^{8} ( 1 + x ^{2} )} d x

\frac{1}{x} - \frac{1}{3 x ^{3}} + \frac{1}{5 x ^{5}} - \frac{1}{7 x ^{7}} + arctan (x) + C

解答ステップ

\int \frac{1}{x ^{8} ( 1 + x ^{2} )} d x

以下の部分分数を得る:

\frac{1}{x ^{8} ( 1 + x ^{2} )} : - \frac{1}{x ^{2}} + \frac{1}{x ^{4}} - \frac{1}{x ^{6}} + \frac{1}{x ^{8}} + \frac{1}{x ^{2} + 1}

= \int - \frac{1}{x ^{2}} + \frac{1}{x ^{4}} - \frac{1}{x ^{6}} + \frac{1}{x ^{8}} + \frac{1}{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}{x ^{2}} d x + \int \frac{1}{x ^{4}} d x - \int \frac{1}{x ^{6}} d x + \int \frac{1}{x ^{8}} d x + \int \frac{1}{x ^{2} + 1} d x

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

\int \frac{1}{x ^{6}} d x = - \frac{1}{5 x ^{5}}

\int \frac{1}{x ^{8}} d x = - \frac{1}{7 x ^{7}}

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

= - (- \frac{1}{x}) - \frac{1}{3 x ^{3}} - (- \frac{1}{5 x ^{5}}) - \frac{1}{7 x ^{7}} + arctan (x)

簡素化

= \frac{1}{x} - \frac{1}{3 x ^{3}} + \frac{1}{5 x ^{5}} - \frac{1}{7 x ^{7}} + arctan (x)

定数を解答に追加する

= \frac{1}{x} - \frac{1}{3 x ^{3}} + \frac{1}{5 x ^{5}} - \frac{1}{7 x ^{7}} + arctan (x) + C