Question

Evaluate the integral $\int{\sqrt{x}\cdot \log (x)dx}$ on the$\left( 0,\infty \right)$.

Hint: Integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative.

We know that, the formula of the integration by parts,
$\int{u\cdot vdx=u\int{vdx-\int{\left[ \dfrac{du}{dx}\int{vdx} \right]dx}}}$, where u and v are the function of x.
When doing Integration by parts, we know that ILATE can be a useful guide most of the time. For those not familiar, ILATE is a guide to help you decide which term to differentiate and which term to integrate.
Where I = Inverse Trigonometric functions,
L = Logarithmic functions,
A = Algebraic functions,
T = Trigonometric functions,
E = Exponential functions
Let $u=\log x$ and $v=\sqrt{x}$ , then the given integral becomes
$\int{\log x\cdot \sqrt{x}dx}=\log x\int{\sqrt{x}dx-\int{\left[ \dfrac{d(\log x)}{dx}\int{\sqrt{x}dx} \right]dx}}..........(1)$
We have $\int{\sqrt{x}dx}=\int{{{x}^{\dfrac{1}{2}}}dx=\dfrac{{{x}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}=\dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}}$
The equation (1) becomes
$\int{\log x\cdot \sqrt{x}dx}=\log x\cdot \dfrac{2}{3}{{x}^{\dfrac{3}{2}}}-\int{\dfrac{1}{x}\cdot \dfrac{2}{3}{{x}^{\dfrac{3}{2}}}dx}$
$\int{\log x\cdot \sqrt{x}dx}=\dfrac{2}{3}\log x\cdot {{x}^{\dfrac{3}{2}}}-\dfrac{2}{3}\int{{{x}^{-1}}\cdot {{x}^{\dfrac{3}{2}}}dx}$
$\int{\log x\cdot \sqrt{x}dx}=\dfrac{2}{3}\log x\cdot {{x}^{\dfrac{3}{2}}}-\dfrac{2}{3}\int{{{x}^{\dfrac{3}{2}-1}}dx}$
$\int{\log x\cdot \sqrt{x}dx}=\dfrac{2}{3}\log x\cdot {{x}^{\dfrac{3}{2}}}-\dfrac{2}{3}\int{{{x}^{\dfrac{1}{2}}}dx}$
$\int{\log x\cdot \sqrt{x}dx}=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}\log x-\dfrac{2}{3}\dfrac{{{x}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}+c$, Where c is the constant of integration
$\int{\log x\cdot \sqrt{x}dx}=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}\log x-\dfrac{2}{3}\dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+c$, Where c is the constant of integration
$\int{\log x\cdot \sqrt{x}dx}=\dfrac{2}{3}{{x}^{\dfrac{3}{2}}}\log x-\dfrac{4}{9}{{x}^{\dfrac{3}{2}}}+c$, Where c is the constant of integration
We have ${{x}^{\dfrac{3}{2}}}=x\sqrt{x}$
$\int{\log x\cdot \sqrt{x}dx}=\dfrac{2}{3}x\sqrt{x}\log x-\dfrac{4}{9}x\sqrt{x}+c$, Where c is the constant of integration
This is the desired result.

Note: When doing Integration by parts, we know that ILATE can be a useful guide most of the time. For those not familiar, ILATE is a guide to help us to decide which term to differentiate and which term to integrate.