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

解答

\int (sin (3 x) - t a g (2 x) + \frac{5}{3 x + 2}) d x

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

求解步骤

\int sin (3 x) - t a g \cdot 2 x + \frac{5}{3 x + 2} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{5}{2 + 3 x} d x + \int sin (3 x) d x - \int 2 a t gx d x

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

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

\int 2 a t gx d x = a t g x^{2}

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

解答补常数

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

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