integral of (sin(3x)-tan(2x)+5/(3x+2))

解

\int (sin (3 x) - tan (2 x) + \frac{5}{3 x + 2}) d x

- \frac{1}{3} cos (3 x) + \frac{1}{2} ln ∣ cos (2 x) ∣ + \frac{5}{3} ln ∣ 3 x + 2 ∣ + C

解答ステップ

\int sin (3 x) - tan (2 x) + \frac{5}{3 x + 2} 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 sin (3 x) d x - \int tan (2 x) d x + \int \frac{5}{3 x + 2} d x

\int sin (3 x) d x = - \frac{1}{3} cos (3 x)

\int tan (2 x) d x = - \frac{1}{2} ln ∣ cos (2 x) ∣

\int \frac{5}{3 x + 2} d x = \frac{5}{3} ln ∣ 3 x + 2 ∣

= - \frac{1}{3} cos (3 x) - (- \frac{1}{2} ln ∣ cos (2 x) ∣) + \frac{5}{3} ln ∣ 3 x + 2 ∣

簡素化

= - \frac{1}{3} cos (3 x) + \frac{1}{2} ln ∣ cos (2 x) ∣ + \frac{5}{3} ln ∣ 3 x + 2 ∣

定数を解答に追加する

= - \frac{1}{3} cos (3 x) + \frac{1}{2} ln ∣ cos (2 x) ∣ + \frac{5}{3} ln ∣ 3 x + 2 ∣ + C