Integral of sqrtx(lnx)dx dx

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

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

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

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

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

      1. Use integration by parts:

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

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

         Then $\operatorname{du}{\left(u \right)} = 2 u$.

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

         1. Let $u = 2 u$.

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

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

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

            Now substitute $u$ back in:

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

         Now evaluate the sub-integral.

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

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

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

         1. Let $u = 2 u$.

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

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

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

            Now substitute $u$ back in:

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

         Now evaluate the sub-integral.

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

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

         1. Let $u = 2 u$.

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

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

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

            Now substitute $u$ back in:

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

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

      So, the result is: $t \left(\frac{u^{2} e^{2 u}}{2} - \frac{u e^{2 u}}{2} + \frac{e^{2 u}}{4}\right)$

   Now substitute $u$ back in:

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

2. Now simplify:

   $$\frac{t x^{2} \left(2 \log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 1\right)}{4}$$

3. Add the constant of integration:

   $$\frac{t x^{2} \left(2 \log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 1\right)}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{t x^{2} \left(2 \log{\left(x \right)}^{2} - 2 \log{\left(x \right)} + 1\right)}{4}+ \mathrm{constant}$$