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Question: Integrate the function \[x{(\log x)^2}\]....

Integrate the function x(log⁡x)2x{(\log x)^2}.

Explanation

Solution

Apply ILATE Method: Integrate in the order from Inverse Logarithmic Algebra Trigonometry and exponential functions, so, integrate by taking logarithmic function as first function and algebraic function as second.

Complete step by step answer :
Given: the given function is x(log⁡x)2x{(\log x)^2}.
Given function with integration sign can be written as ∫x(log⁡x)2dx\int {x{{(\log x)}^2}dx} .
Apply the ILATE method for integration of the provided expression.
To solve the given expression, use the formula:
first function∫second function−∫(ddxfirst function∫second functiondx)dx{\text{first function}}\int {{\text{second function}} - \int {\left( {\dfrac{d}{{dx}}{\text{first function}}\int {{\text{second functiondx}}} } \right)dx} } .
There are two types of functions in the given function, first is of algebraic type and second in the parantheses is of logarithmic type.
By using ILATE rule, take x as the second function and (log⁡x)2{\left( {\log x} \right)^2} as the first function and substitute in the below formula:
first function∫second function−∫(ddxfirst function∫second functiondx)dx{\text{first function}}\int {{\text{second function}} - \int {\left( {\dfrac{d}{{dx}}{\text{first function}}\int {{\text{second functiondx}}} } \right)dx} }
(log⁡x)2∫xdx−∫(ddx(log⁡x)2∫xdx)dx{(\log x)^2}\int {xdx - } \int {\left( {\dfrac{d}{{dx}}{{\left( {\log x} \right)}^2}\int {xdx} } \right)} dx
It is known that ddx(log⁡x)=1x, ddx(xn)=nxn−1\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x},{\text{ }}\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} and ∫xndx=xn+1n+1+C\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} .
So,

(log⁡x)2(x22)−∫(2log⁡x(1x)(x22))dx (log⁡x)2(x22)−∫xlog⁡xdx  {(\log x)^2}\left( {\dfrac{{{x^2}}}{2}} \right) - \int {\left( {2\log x\left( {\dfrac{1}{x}} \right)\left( {\dfrac{{{x^2}}}{2}} \right)} \right)dx} \\\ {(\log x)^2}\left( {\dfrac{{{x^2}}}{2}} \right) - \int {x\log xdx} \\\

To solve the second integration term, take x as the second function and log⁡x\log x as the first function and substitute in the below formula:
first function∫second function−∫(ddxfirst function∫second functiondx)dx{\text{first function}}\int {{\text{second function}} - \int {\left( {\dfrac{d}{{dx}}{\text{first function}}\int {{\text{second functiondx}}} } \right)dx} }
(log⁡x)2(x22)−[log⁡x∫xdx−∫(ddx(log⁡x)∫xdx)dx]{(\log x)^2}\left( {\dfrac{{{x^2}}}{2}} \right) - \left[ {\log x\int {xdx - \int {\left( {\dfrac{d}{{dx}}(\log x)\int {xdx} } \right)dx} } } \right]
Again use the formula of integration and differentiation, ddx(log⁡x)=1x, ddx(xn)=nxn−1\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x},{\text{ }}\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} and ∫xndx=xn+1n+1+C\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} .

(log⁡x)2(x22)−[log⁡x(x22)−∫1x×x22dx] (log⁡x)2(x22)−[log⁡x(x22)−∫x2dx]  {(\log x)^2}\left( {\dfrac{{{x^2}}}{2}} \right) - \left[ {\log x\left( {\dfrac{{{x^2}}}{2}} \right) - \int {\dfrac{1}{x} \times \dfrac{{{x^2}}}{2}dx} } \right] \\\ {(\log x)^2}\left( {\dfrac{{{x^2}}}{2}} \right) - \left[ {\log x\left( {\dfrac{{{x^2}}}{2}} \right) - \int {\dfrac{x}{2}dx} } \right] \\\

Simplify the obtained expression further by using ∫xndx=xn+1n+1+C\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} .

(log⁡x)2(x22)−[log⁡x(x22)−12×x22+C] (log⁡x)2(x22)−[log⁡x(x22)−x24+C] (log⁡x)2(x22)−log⁡x(x22)+x24+C  {(\log x)^2}\left( {\dfrac{{{x^2}}}{2}} \right) - \left[ {\log x\left( {\dfrac{{{x^2}}}{2}} \right) - \dfrac{1}{2} \times \dfrac{{{x^2}}}{2} + C} \right] \\\ {(\log x)^2}\left( {\dfrac{{{x^2}}}{2}} \right) - \left[ {\log x\left( {\dfrac{{{x^2}}}{2}} \right) - \dfrac{{{x^2}}}{4} + C} \right] \\\ {(\log x)^2}\left( {\dfrac{{{x^2}}}{2}} \right) - \log x\left( {\dfrac{{{x^2}}}{2}} \right) + \dfrac{{{x^2}}}{4} + C \\\

Hence, ∫x(log⁡x)2dx=(log⁡x)2(x22)−log⁡x(x22)+x24+C\int {x{{(\log x)}^2}dx} = {(\log x)^2}\left( {\dfrac{{{x^2}}}{2}} \right) - \log x\left( {\dfrac{{{x^2}}}{2}} \right) + \dfrac{{{x^2}}}{4} + C, or it can be said that the integration of the function x(log⁡x)2x{(\log x)^2} is (log⁡x)2(x22)−log⁡x(x22)+x24+C{(\log x)^2}\left( {\dfrac{{{x^2}}}{2}} \right) - \log x\left( {\dfrac{{{x^2}}}{2}} \right) + \dfrac{{{x^2}}}{4} + C.

Note: ILATE method is applied two times to find the integration easily. In such a problem, any other method applied to find the integration will make it more complicated.