integral of 1/(1+\frac{1){x^2}}

解

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

- arctan (x) + x + C

解答ステップ

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

簡素化

\frac{1}{1 + \frac{1}{x ^{2}}} : \frac{x ^{2}}{x ^{2} + 1}

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

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

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

\int \frac{1}{x ^{2} + 1} d x = arctan (x)

\int 1 d x = x

= - arctan (x) + x

定数を解答に追加する

= - arctan (x) + x + C