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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.

__Complete step-by-step answer:__

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.

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.

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