Integral of ln(x^2)/x dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \log{\left(x^{2} \right)}$.

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

      $$\int \frac{u}{4}\, du$$

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

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

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

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

         So, the result is: $\frac{u^{2}}{4}$

      Now substitute $u$ back in:

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

   Method #2

   1. Let $u = x^{2}$.

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

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

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

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

         1. Let $u = \frac{1}{u}$.

            Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$:

            $$\int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du$$

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

               $$\int \left(- \frac{\log{\left(\frac{1}{u} \right)}}{u}\right)\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du$$

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

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

                  $$\int u\, du$$

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

                     $$\int \left(- u\right)\, du = - \int u\, du$$

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

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

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

                  Now substitute $u$ back in:

                  $$- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}$$

               So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}$

            Now substitute $u$ back in:

            $$\frac{\log{\left(u \right)}^{2}}{2}$$

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

      Now substitute $u$ back in:

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

   Method #3

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

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

      $$\int \frac{\log{\left(\frac{1}{u^{2}} \right)}}{u}\, du$$

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

         $$\int \left(- \frac{\log{\left(\frac{1}{u^{2}} \right)}}{u}\right)\, du = - \int \frac{\log{\left(\frac{1}{u^{2}} \right)}}{u}\, du$$

         1. Let $u = \log{\left(\frac{1}{u^{2}} \right)}$.

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

            $$\int \frac{u}{4}\, du$$

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

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

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

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

               So, the result is: $- \frac{u^{2}}{4}$

            Now substitute $u$ back in:

            $$- \frac{\log{\left(\frac{1}{u^{2}} \right)}^{2}}{4}$$

         So, the result is: $\frac{\log{\left(\frac{1}{u^{2}} \right)}^{2}}{4}$

      Now substitute $u$ back in:

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

2. Add the constant of integration:

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

The answer is:

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