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

解

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

\frac{x ^{2}}{2} - 2 x + \frac{1}{8} ln ∣ x ∣ + \frac{1}{4 x} - \frac{1}{4 x ^{2}} + \frac{31}{8} ln ∣ x + 2 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{x ^{5} + 1}{x ^{3} ( x + 2 )} : x - 2 + \frac{1}{8 x} - \frac{1}{4 x ^{2}} + \frac{1}{2 x ^{3}} + \frac{31}{8 ( x + 2 )}

= \int x - 2 + \frac{1}{8 x} - \frac{1}{4 x ^{2}} + \frac{1}{2 x ^{3}} + \frac{31}{8 ( x + 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 x d x - \int 2 d x + \int \frac{1}{8 x} d x - \int \frac{1}{4 x ^{2}} d x + \int \frac{1}{2 x ^{3}} d x + \int \frac{31}{8 ( x + 2 )} d x

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

\int 2 d x = 2 x

\int \frac{1}{8 x} d x = \frac{1}{8} ln ∣ x ∣

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

\int \frac{1}{2 x ^{3}} d x = - \frac{1}{4 x ^{2}}

\int \frac{31}{8 ( x + 2 )} d x = \frac{31}{8} ln ∣ x + 2 ∣

= \frac{x ^{2}}{2} - 2 x + \frac{1}{8} ln ∣ x ∣ - (- \frac{1}{4 x}) - \frac{1}{4 x ^{2}} + \frac{31}{8} ln ∣ x + 2 ∣

簡素化

= \frac{x ^{2}}{2} - 2 x + \frac{1}{8} ln ∣ x ∣ + \frac{1}{4 x} - \frac{1}{4 x ^{2}} + \frac{31}{8} ln ∣ x + 2 ∣

定数を解答に追加する

= \frac{x ^{2}}{2} - 2 x + \frac{1}{8} ln ∣ x ∣ + \frac{1}{4 x} - \frac{1}{4 x ^{2}} + \frac{31}{8} ln ∣ x + 2 ∣ + C