integral of 1/(sqrt(x))+4/(x^2)-1

解

\int \frac{1}{x} + \frac{4}{x ^{2}} - 1 d x

2 x - \frac{4}{x} - x + C

解答ステップ

\int \frac{1}{x} + \frac{4}{x ^{2}} - 1 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{1}{x} d x + \int \frac{4}{x ^{2}} d x - \int 1 d x

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

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

\int 1 d x = x

= 2 x - \frac{4}{x} - x

定数を解答に追加する

= 2 x - \frac{4}{x} - x + C