Integral of (x^3)lnx dx

1. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int u e^{4 u}\, du$$

      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 $\operatorname{dv}{\left(u \right)} = e^{4 u}$.

         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 $u = 4 u$.

               Then let $du = 4 du$ and substitute $\frac{du}{4}$:

               $$\int \frac{e^{u}}{16}\, du$$

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

                  $$\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$$

                  1. The integral of the exponential function is itself.

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

                  So, the result is: $\frac{e^{u}}{4}$

               Now substitute $u$ back in:

               $$\frac{e^{4 u}}{4}$$

            Method #2

            1. Let $u = e^{4 u}$.

               Then let $du = 4 e^{4 u} du$ and substitute $\frac{du}{4}$:

               $$\int \frac{1}{16}\, du$$

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

                  $$\int \frac{1}{4}\, du = \frac{\int 1\, du}{4}$$

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

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

                  So, the result is: $\frac{u}{4}$

               Now substitute $u$ back in:

               $$\frac{e^{4 u}}{4}$$

         Now evaluate the sub-integral.

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

         $$\int \frac{e^{4 u}}{4}\, du = \frac{\int e^{4 u}\, du}{4}$$

         1. Let $u = 4 u$.

            Then let $du = 4 du$ and substitute $\frac{du}{4}$:

            $$\int \frac{e^{u}}{16}\, du$$

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

               $$\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}$$

               1. The integral of the exponential function is itself.

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

               So, the result is: $\frac{e^{u}}{4}$

            Now substitute $u$ back in:

            $$\frac{e^{4 u}}{4}$$

         So, the result is: $\frac{e^{4 u}}{16}$

      Now substitute $u$ back in:

      $$\frac{x^{4} \log{\left(x \right)}}{4} - \frac{x^{4}}{16}$$

   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 $\operatorname{dv}{\left(x \right)} = x^{3}$.

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

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

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

         $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

      Now evaluate the sub-integral.

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

      $$\int \frac{x^{3}}{4}\, dx = \frac{\int x^{3}\, dx}{4}$$

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

         $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

      So, the result is: $\frac{x^{4}}{16}$

2. Now simplify:

   $$\frac{x^{4} \cdot \left(4 \log{\left(x \right)} - 1\right)}{16}$$

3. Add the constant of integration:

   $$\frac{x^{4} \cdot \left(4 \log{\left(x \right)} - 1\right)}{16}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{4} \cdot \left(4 \log{\left(x \right)} - 1\right)}{16}+ \mathrm{constant}$$