Integral of y^2lnydy dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \log{\left(y \right)}$.

      Then let $du = \frac{dy}{y}$ and substitute $du$:

      $$\int u e^{3 u}\, du$$

      1. Use integration by parts:

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

         Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{3 u}$.

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

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

         1. There are multiple ways to do this integral.

            Method #1

            1. Let $u = 3 u$.

               Then let $du = 3 du$ and substitute $\frac{du}{3}$:

               $$\int \frac{e^{u}}{9}\, du$$

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

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

                  1. The integral of the exponential function is itself.

                     $$\int e^{u}\, du = e^{u}$$

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

               Now substitute $u$ back in:

               $$\frac{e^{3 u}}{3}$$

            Method #2

            1. Let $u = e^{3 u}$.

               Then let $du = 3 e^{3 u} du$ and substitute $\frac{du}{3}$:

               $$\int \frac{1}{9}\, du$$

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

                  $$\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$$

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

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

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

               Now substitute $u$ back in:

               $$\frac{e^{3 u}}{3}$$

         Now evaluate the sub-integral.

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

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

         1. Let $u = 3 u$.

            Then let $du = 3 du$ and substitute $\frac{du}{3}$:

            $$\int \frac{e^{u}}{9}\, du$$

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

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

               1. The integral of the exponential function is itself.

                  $$\int e^{u}\, du = e^{u}$$

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

            Now substitute $u$ back in:

            $$\frac{e^{3 u}}{3}$$

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

      Now substitute $u$ back in:

      $$\frac{y^{3} \log{\left(y \right)}}{3} - \frac{y^{3}}{9}$$

   Method #2

   1. Use integration by parts:

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

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

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

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

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

         $$\int y^{2}\, dy = \frac{y^{3}}{3}$$

      Now evaluate the sub-integral.

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

      $$\int \frac{y^{2}}{3}\, dy = \frac{\int y^{2}\, dy}{3}$$

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

         $$\int y^{2}\, dy = \frac{y^{3}}{3}$$

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

2. Now simplify:

   $$\frac{y^{3} \cdot \left(3 \log{\left(y \right)} - 1\right)}{9}$$

3. Add the constant of integration:

   $$\frac{y^{3} \cdot \left(3 \log{\left(y \right)} - 1\right)}{9}+ \mathrm{constant}$$

The answer is:

$$\frac{y^{3} \cdot \left(3 \log{\left(y \right)} - 1\right)}{9}+ \mathrm{constant}$$