integral of (1-x^2)/(2x)

解

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

\frac{1}{2} (ln ∣ x ∣ - \frac{x ^{2}}{2}) + C

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

拡張

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

= \frac{1}{2} \cdot \int \frac{1}{x} - x 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

= \frac{1}{2} (\int \frac{1}{x} d x - \int x d x)

\int \frac{1}{x} d x = ln ∣ x ∣

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

= \frac{1}{2} (ln ∣ x ∣ - \frac{x ^{2}}{2})

定数を解答に追加する

= \frac{1}{2} (ln ∣ x ∣ - \frac{x ^{2}}{2}) + C