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

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

Explanation

Solution

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,
∫u⋅vdx=u∫vdx−∫[dudx∫vdx]dx\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⁡xu=\log x and v=xv=\sqrt{x} , then the given integral becomes
∫log⁡x⋅xdx=log⁡x∫xdx−∫[d(log⁡x)dx∫xdx]dx..........(1)\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 ∫xdx=∫x12dx=x12+112+1=x3232=23x32\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
∫log⁡x⋅xdx=log⁡x⋅23x32−∫1x⋅23x32dx\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}
∫log⁡x⋅xdx=23log⁡x⋅x32−23∫x−1⋅x32dx\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}
∫log⁡x⋅xdx=23log⁡x⋅x32−23∫x32−1dx\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}
∫log⁡x⋅xdx=23log⁡x⋅x32−23∫x12dx\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}
∫log⁡x⋅xdx=23x32log⁡x−23x12+112+1+c\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
∫log⁡x⋅xdx=23x32log⁡x−23x3232+c\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
∫log⁡x⋅xdx=23x32log⁡x−49x32+c\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 x32=xx{{x}^{\dfrac{3}{2}}}=x\sqrt{x}
∫log⁡x⋅xdx=23xxlog⁡x−49xx+c\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.