integral of 1/(x^5+16x^3+64x)

解

\int \frac{1}{x ^{5} + 16 x ^{3} + 64 x} d x

\frac{1}{64} ln ∣ x ∣ - \frac{1}{128} ln x^{2} + 8 + \frac{1}{16 ( x ^{2} + 8 )} + C

解答ステップ

\int \frac{1}{x ^{5} + 16 x ^{3} + 64 x} d x

以下の部分分数を得る:

\frac{1}{x ^{5} + 16 x ^{3} + 64 x} : \frac{1}{64 x} - \frac{x}{64 ( x ^{2} + 8 )} - \frac{x}{8 ( x ^{2} + 8 ) ^{2}}

= \int \frac{1}{64 x} - \frac{x}{64 ( x ^{2} + 8 )} - \frac{x}{8 ( x ^{2} + 8 ) ^{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}{64 x} d x - \int \frac{x}{64 ( x ^{2} + 8 )} d x - \int \frac{x}{8 ( x ^{2} + 8 ) ^{2}} d x

\int \frac{1}{64 x} d x = \frac{1}{64} ln ∣ x ∣

\int \frac{x}{64 ( x ^{2} + 8 )} d x = \frac{1}{128} ln x^{2} + 8

\int \frac{x}{8 ( x ^{2} + 8 ) ^{2}} d x = - \frac{1}{16 ( x ^{2} + 8 )}

= \frac{1}{64} ln ∣ x ∣ - \frac{1}{128} ln x^{2} + 8 - (- \frac{1}{16 ( x ^{2} + 8 )})

簡素化

= \frac{1}{64} ln ∣ x ∣ - \frac{1}{128} ln x^{2} + 8 + \frac{1}{16 ( x ^{2} + 8 )}

定数を解答に追加する

= \frac{1}{64} ln ∣ x ∣ - \frac{1}{128} ln x^{2} + 8 + \frac{1}{16 ( x ^{2} + 8 )} + C