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

解

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

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

解答ステップ

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

\int \frac{3}{x ^{4}} d x = - \frac{1}{x ^{3}}

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

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

= - \frac{1}{x ^{3}} - \frac{4}{x} - (- \frac{3}{x ^{2}})

簡素化

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

定数を解答に追加する

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