Integral of ln(x)/√x 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 $du$:

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

      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 $\operatorname{dv}{\left(u \right)} = e^{\frac{u}{2}}$.

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

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

         1. Let $u = \frac{u}{2}$.

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

            $$\int 2 e^{u}\, du$$

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

               $$\text{False}$$

               1. The integral of the exponential function is itself.

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

               So, the result is: $2 e^{u}$

            Now substitute $u$ back in:

            $$2 e^{\frac{u}{2}}$$

         Now evaluate the sub-integral.

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

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

         1. Let $u = \frac{u}{2}$.

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

            $$\int 2 e^{u}\, du$$

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

               $$\text{False}$$

               1. The integral of the exponential function is itself.

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

               So, the result is: $2 e^{u}$

            Now substitute $u$ back in:

            $$2 e^{\frac{u}{2}}$$

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

      Now substitute $u$ back in:

      $$2 \sqrt{x} \log{\left(x \right)} - 4 \sqrt{x}$$

   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 $\operatorname{dv}{\left(x \right)} = \frac{1}{\sqrt{x}}$.

      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 $n \neq -1$:

         $$\int \frac{1}{\sqrt{x}}\, dx = 2 \sqrt{x}$$

      Now evaluate the sub-integral.

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

      $$\int \frac{2}{\sqrt{x}}\, dx = 2 \int \frac{1}{\sqrt{x}}\, dx$$

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

         $$\int \frac{1}{\sqrt{x}}\, dx = 2 \sqrt{x}$$

      So, the result is: $4 \sqrt{x}$

2. Now simplify:

   $$2 \sqrt{x} \left(\log{\left(x \right)} - 2\right)$$

3. Add the constant of integration:

   $$2 \sqrt{x} \left(\log{\left(x \right)} - 2\right)+ \mathrm{constant}$$

The answer is:

$$2 \sqrt{x} \left(\log{\left(x \right)} - 2\right)+ \mathrm{constant}$$