integral of [cot(x)+1/x ]

解

\int [cot (x) + \frac{1}{x}] d x

ln ∣ sin (x) ∣ + ln ∣ x ∣ + C

解答ステップ

\int cot (x) + \frac{1}{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

= \int cot (x) d x + \int \frac{1}{x} d x

\int cot (x) d x = ln ∣ sin (x) ∣

\int \frac{1}{x} d x = ln ∣ x ∣

= ln ∣ sin (x) ∣ + ln ∣ x ∣

定数を解答に追加する

= ln ∣ sin (x) ∣ + ln ∣ x ∣ + C