# Find the indefinite integral: \int -\frac{2}{x} + 3x + \sin(2x) dx

## Question:

Find the indefinite integral:

{eq}\displaystyle \int -\frac{2}{x} + 3x + \sin(2x)\;dx {/eq}

## Integration By Substitution:

To find the solution to the indefinite integral, use the sum rule and then apply the u-substitution method. The u-substitution method is also known as integration by substitution. The sum rule is {eq}{\color{Blue}{\displaystyle \int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx}} {/eq}.

{eq}\displaystyle \int -\frac{2}{x}+3x+\sin (2x)dx = -\int \frac{2}{3}dx + \int 3xdx + \int \sin(2x) dx \ \ \ \ \ {\color{Blue}{\text{(Apply the sum rule:} \displaystyle \int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx)}} {/eq}

Consider:

{eq}\begin{align*} -\int \frac{2}{3}dx &= \frac{2}{3}\int dx &\text{(Take the constant out)} \\ &= \frac{2}{3}x \end{align*} {/eq}

Consider:

{eq}\begin{align*} \int 3xdx + \int \sin(2x) dx &= 3 \int x dx &\text{(Take the constant out)} \\ &= 3 \frac{x^2}{2} &{\color{Blue}{\text{(Apply the power rule:} \displaystyle \int x^n dx = \frac{x^{n + 1}}{n + 1}, where \ n \neq -1)}} \end{align*} {/eq}

Consider:

{eq}\displaystyle \int \sin(2x) dx {/eq}

Apply u-substitution:

{eq}u = 2x \\ du = 2 dx {/eq}

{eq}\begin{align*} \int \sin(2x) dx &= \int \sin(u) \frac{du}{2} \\ &= \frac{1}{2}\int \sin(u) du &\text{(Take the constant out)}\\ &= \frac{1}{2}(-\cos(u)) \\ &= - \frac{1}{2}\cos(2x) &\text{(Where,}\ u = 2x) \end{align*} {/eq}

Therefore, {eq}\displaystyle \int -\frac{2}{x}+3x+\sin (2x)dx = - \frac{2}{3}x + 3 \frac{x^2}{2} - \frac{1}{2}\cos(2x) +C {/eq}

Therefore, the solution is {eq}\boxed{\displaystyle {\color{Blue}{\int -\frac{2}{x}+3x+\sin (2x)dx = - \frac{2}{3}x + 3 \frac{x^2}{2} - \frac{1}{2}\cos(2x) +C}}} {/eq}