Integral of ln(e^x) dx

1. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = e^{x} dx$ and substitute $du$:

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

      1. There are multiple ways to do this integral.

         Method #1

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

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

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

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

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

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

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

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

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

                     $$\int u\, du = - \int u\, du$$

                     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:

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

         Method #2

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

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

            $$\int u\, du$$

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

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

            Now substitute $u$ back in:

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

      Now substitute $u$ back in:

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

   Method #2

   1. Use integration by parts:

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

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

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

      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 $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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