Integral of 3sqrtxlnx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \sqrt{x}$.

      Then let $du = \frac{dx}{2 \sqrt{x}}$ and substitute $6 du$:

      $$\int 6 u^{2} \log{\left(u^{2} \right)}\, du$$

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

         $$\int u^{2} \log{\left(u^{2} \right)}\, du = 6 \int u^{2} \log{\left(u^{2} \right)}\, du$$

         1. Use integration by parts:

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

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

            Then $\operatorname{du}{\left(u \right)} = \frac{2}{u}$.

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

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

               $$\int u^{2}\, du = \frac{u^{3}}{3}$$

            Now evaluate the sub-integral.

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

            $$\int \frac{2 u^{2}}{3}\, du = \frac{2 \int u^{2}\, du}{3}$$

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

               $$\int u^{2}\, du = \frac{u^{3}}{3}$$

            So, the result is: $\frac{2 u^{3}}{9}$

         So, the result is: $2 u^{3} \log{\left(u^{2} \right)} - \frac{4 u^{3}}{3}$

      Now substitute $u$ back in:

      $$2 x^{\frac{3}{2}} \log{\left(x \right)} - \frac{4 x^{\frac{3}{2}}}{3}$$

   Method #2

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 3 \sqrt{x}$.

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

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

      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: $2 x^{\frac{3}{2}}$

      Now evaluate the sub-integral.

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

      $$\int 2 \sqrt{x}\, dx = 2 \int \sqrt{x}\, dx$$

      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: $\frac{4 x^{\frac{3}{2}}}{3}$

2. Now simplify:

   $$x^{\frac{3}{2}} \left(2 \log{\left(x \right)} - \frac{4}{3}\right)$$

3. Add the constant of integration:

   $$x^{\frac{3}{2}} \left(2 \log{\left(x \right)} - \frac{4}{3}\right)+ \mathrm{constant}$$

The answer is:

$$x^{\frac{3}{2}} \left(2 \log{\left(x \right)} - \frac{4}{3}\right)+ \mathrm{constant}$$