integral of (x^4+1)/(x^3+9x)

解

\int \frac{x ^{4} + 1}{x ^{3} + 9 x} d x

\frac{x ^{2}}{2} + \frac{1}{9} ln ∣ x ∣ - \frac{41}{9} ln x^{2} + 9 + C

解答ステップ

\int \frac{x ^{4} + 1}{x ^{3} + 9 x} d x

以下の部分分数を得る:

\frac{x ^{4} + 1}{x ^{3} + 9 x} : x + \frac{1}{9 x} - \frac{82 x}{9 ( x ^{2} + 9 )}

= \int x + \frac{1}{9 x} - \frac{82 x}{9 ( x ^{2} + 9 )} 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 x d x + \int \frac{1}{9 x} d x - \int \frac{82 x}{9 ( x ^{2} + 9 )} d x

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

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

\int \frac{82 x}{9 ( x ^{2} + 9 )} d x = \frac{41}{9} ln x^{2} + 9

= \frac{x ^{2}}{2} + \frac{1}{9} ln ∣ x ∣ - \frac{41}{9} ln x^{2} + 9

定数を解答に追加する

= \frac{x ^{2}}{2} + \frac{1}{9} ln ∣ x ∣ - \frac{41}{9} ln x^{2} + 9 + C