integral of (x^2+19x+10)/(2x^4+5x^3)

解

\int \frac{x ^{2} + 19 x + 10}{2 x ^{4} + 5 x ^{3}} d x

- ln ∣ x ∣ - \frac{3}{x} - \frac{1}{x ^{2}} + ln ∣ 2 x + 5 ∣ + C

解答ステップ

\int \frac{x ^{2} + 19 x + 10}{2 x ^{4} + 5 x ^{3}} d x

以下の部分分数を得る:

\frac{x ^{2} + 19 x + 10}{2 x ^{4} + 5 x ^{3}} : - \frac{1}{x} + \frac{3}{x ^{2}} + \frac{2}{x ^{3}} + \frac{2}{2 x + 5}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

\int \frac{2}{2 x + 5} d x = ln ∣ 2 x + 5 ∣

= - ln ∣ x ∣ - \frac{3}{x} - \frac{1}{x ^{2}} + ln ∣ 2 x + 5 ∣

定数を解答に追加する

= - ln ∣ x ∣ - \frac{3}{x} - \frac{1}{x ^{2}} + ln ∣ 2 x + 5 ∣ + C