integral of (3x^4-x^3+6x^2)/(x^4)

解

\int \frac{3 x ^{4} - x ^{3} + 6 x ^{2}}{x ^{4}} d x

3 x - ln ∣ x ∣ - \frac{6}{x} + C

解答ステップ

\int \frac{3 x ^{4} - x ^{3} + 6 x ^{2}}{x ^{4}} d x

拡張

\frac{3 x ^{4} - x ^{3} + 6 x ^{2}}{x ^{4}} : 3 - \frac{1}{x} + \frac{6}{x ^{2}}

= \int 3 - \frac{1}{x} + \frac{6}{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 3 d x - \int \frac{1}{x} d x + \int \frac{6}{x ^{2}} d x

\int 3 d x = 3 x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{6}{x ^{2}} d x = - \frac{6}{x}

= 3 x - ln ∣ x ∣ - \frac{6}{x}

定数を解答に追加する

= 3 x - ln ∣ x ∣ - \frac{6}{x} + C