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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

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

簡素化

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

定数を解答に追加する

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