Integral of (log2x+3×sqrt(x)) dx

1. Integrate term-by-term:

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int 3 \sqrt{x}\, dx = 3 \int \sqrt{x}\, dx$$

      1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int \sqrt{x}\, dx = \frac{2 x^{\frac{3}{2}}}{3}$$

      So, the result is: $2 x^{\frac{3}{2}}$

   1. There are multiple ways to do this integral.

      Method #1

      1. Let $u = 2 x$.

         Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

         $$\int \frac{\log{\left(u \right)}}{4}\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \frac{\log{\left(u \right)}}{2}\, du = \frac{\int \log{\left(u \right)}\, du}{2}$$

            1. Use integration by parts:

               $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

               Let $u{\left(u \right)} = \log{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = 1$.

               Then $\operatorname{du}{\left(u \right)} = \frac{1}{u}$.

               To find $v{\left(u \right)}$:

               1. The integral of a constant is the constant times the variable of integration:

                  $$\int 1\, du = u$$

               Now evaluate the sub-integral.

            2. The integral of a constant is the constant times the variable of integration:

               $$\int 1\, du = u$$

            So, the result is: $\frac{u \log{\left(u \right)}}{2} - \frac{u}{2}$

         Now substitute $u$ back in:

         $$x \log{\left(2 x \right)} - x$$

      Method #2

      1. Use integration by parts:

         $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

         Let $u{\left(x \right)} = \log{\left(2 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

         Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$.

         To find $v{\left(x \right)}$:

         1. The integral of a constant is the constant times the variable of integration:

            $$\int 1\, dx = x$$

         Now evaluate the sub-integral.

      2. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, dx = x$$

   The result is: $2 x^{\frac{3}{2}} + x \log{\left(2 x \right)} - x$

2. Add the constant of integration:

   $$2 x^{\frac{3}{2}} + x \log{\left(2 x \right)} - x+ \mathrm{constant}$$

The answer is:

$$2 x^{\frac{3}{2}} + x \log{\left(2 x \right)} - x+ \mathrm{constant}$$