What is the integral of (sin(3x)-tag(2x)+5/(3x+2)) ?

The integral of (sin(3x)-tag(2x)+5/(3x+2)) is 5/3 ln|2+3x|-1/3 cos(3x)-atgx^2+C

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

\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

Solution steps

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

Apply the Sum Rule:

\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}

Add a constant to the solution

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

\int 2 - 4 t + 5 t^{\frac{3}{2}} d t

\int \frac{x - x ^{3} \cdot e ^{x} + x ^{2}}{x ^{3}} d x

\int (5 - \frac{1}{x}) d x

\int sin^{5} (2 x) cos^{5} (2 x) d x

\int \frac{1}{4 + 6 x - x ^{2}} d x

What is the integral of (sin(3x)-tag(2x)+5/(3x+2)) ?
The integral of (sin(3x)-tag(2x)+5/(3x+2)) is 5/3 ln|2+3x|-1/3 cos(3x)-atgx^2+C