integral of (x^4-3x^2+x-sqrt(x)+2)/x

解

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

\frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} + x - 2 x + 2 ln ∣ x ∣ + C

解答ステップ

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

拡張

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

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

\int x^{3} d x = \frac{x ^{4}}{4}

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

\int 1 d x = x

\int \frac{1}{x} d x = 2 x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

= \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} + x - 2 x + 2 ln ∣ x ∣

定数を解答に追加する

= \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2} + x - 2 x + 2 ln ∣ x ∣ + C