integral of (x^5+1)/(x^3-3x^2-10x)

解

\int \frac{x ^{5} + 1}{x ^{3} - 3 x ^{2} - 10 x} d x

\frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} + 19 x - \frac{1}{10} ln ∣ x ∣ - \frac{31}{14} ln ∣ x + 2 ∣ + \frac{3126}{35} ln ∣ x - 5 ∣ + C

解答ステップ

\int \frac{x ^{5} + 1}{x ^{3} - 3 x ^{2} - 10 x} d x

以下の部分分数を得る:

\frac{x ^{5} + 1}{x ^{3} - 3 x ^{2} - 10 x} : x^{2} + 3 x + 19 - \frac{1}{10 x} - \frac{31}{14 ( x + 2 )} + \frac{3126}{35 ( x - 5 )}

= \int x^{2} + 3 x + 19 - \frac{1}{10 x} - \frac{31}{14 ( x + 2 )} + \frac{3126}{35 ( x - 5 )} 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^{2} d x + \int 3 x d x + \int 19 d x - \int \frac{1}{10 x} d x - \int \frac{31}{14 ( x + 2 )} d x + \int \frac{3126}{35 ( x - 5 )} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 3 x d x = \frac{3 x ^{2}}{2}

\int 19 d x = 19 x

\int \frac{1}{10 x} d x = \frac{1}{10} ln ∣ x ∣

\int \frac{31}{14 ( x + 2 )} d x = \frac{31}{14} ln ∣ x + 2 ∣

\int \frac{3126}{35 ( x - 5 )} d x = \frac{3126}{35} ln ∣ x - 5 ∣

= \frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} + 19 x - \frac{1}{10} ln ∣ x ∣ - \frac{31}{14} ln ∣ x + 2 ∣ + \frac{3126}{35} ln ∣ x - 5 ∣

定数を解答に追加する

= \frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} + 19 x - \frac{1}{10} ln ∣ x ∣ - \frac{31}{14} ln ∣ x + 2 ∣ + \frac{3126}{35} ln ∣ x - 5 ∣ + C