integral of (3x^6+4x^2+12)/(x^4)

解

\int \frac{3 x ^{6} + 4 x ^{2} + 12}{x ^{4}} d x

x^{3} - \frac{4}{x} - \frac{4}{x ^{3}} + C

解答ステップ

\int \frac{3 x ^{6} + 4 x ^{2} + 12}{x ^{4}} d x

拡張

\frac{3 x ^{6} + 4 x ^{2} + 12}{x ^{4}} : 3 x^{2} + \frac{4}{x ^{2}} + \frac{12}{x ^{4}}

= \int 3 x^{2} + \frac{4}{x ^{2}} + \frac{12}{x ^{4}} 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 x^{2} d x + \int \frac{4}{x ^{2}} d x + \int \frac{12}{x ^{4}} d x

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

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

\int \frac{12}{x ^{4}} d x = - \frac{4}{x ^{3}}

= x^{3} - \frac{4}{x} - \frac{4}{x ^{3}}

定数を解答に追加する

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