Question

Solve the integral:$\int x\ln \sqrt{x}dx$

Short answer

The required answer is$\frac{x^2}{4}\ln \left(x\right)-\frac{x^2}{8}+c$.

Step-by-step solution

Step 1. Given information. 

We have given integral is$\int x\ln \sqrt{x}dx$.

Step 2. Solve the integration by parts .  

We have,

$$\begin{gathered} \int x\ln \sqrt{x}dx \\ =\int \left(\frac{x}{2}\ln x\right)dx \\ =\frac{1}{2}\int x\ln xdx \end{gathered}$$

$$\begin{gathered} u=\ln x \\ du=\frac{dx}{x} \end{gathered}$$

and

$$\begin{gathered} dv=xdx \\ v=\int xdx \\ v=\frac{x^2}{2} \end{gathered}$$

The formula of integration by parts is $\int udv=uv-\int vdu$.

localid="1648797136253" $$\begin{gathered} \frac{1}{2}\int x\ln xdx \\ =\frac{1}{2}\left(\ln \left(x\right)\frac{x^2}{2}-\int \frac{x^2}{2}\frac{1}{x}dx\right) \\ =\frac{1}{2}\left(\frac{x^2}{2}\ln \left(x\right)-\frac{1}{2}\int xdx\right) \\ =\frac{x^2}{4}\ln \left(x\right)-\frac{x^2}{8}+c \end{gathered}$$