integral of ((2x^4-8x^3-6x^2+4)/(x^3))

解

\int (\frac{2 x ^{4} - 8 x ^{3} - 6 x ^{2} + 4}{x ^{3}}) d x

x^{2} - 8 x - 6 ln ∣ x ∣ - \frac{2}{x ^{2}} + C

解答ステップ

\int \frac{2 x ^{4} - 8 x ^{3} - 6 x ^{2} + 4}{x ^{3}} d x

拡張

\frac{2 x ^{4} - 8 x ^{3} - 6 x ^{2} + 4}{x ^{3}} : 2 x - 8 - \frac{6}{x} + \frac{4}{x ^{3}}

= \int 2 x - 8 - \frac{6}{x} + \frac{4}{x ^{3}} 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 2 x d x - \int 8 d x - \int \frac{6}{x} d x + \int \frac{4}{x ^{3}} d x

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

\int 8 d x = 8 x

\int \frac{6}{x} d x = 6 ln ∣ x ∣

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

= x^{2} - 8 x - 6 ln ∣ x ∣ - \frac{2}{x ^{2}}

定数を解答に追加する

= x^{2} - 8 x - 6 ln ∣ x ∣ - \frac{2}{x ^{2}} + C