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Question: How do you integrate \(x\ln {\left( x \right)^2}\)?...

How do you integrate xln⁡(x)2x\ln {\left( x \right)^2}?

Explanation

Solution

Given the expression. We have to find the integral of the expression by applying integration by parts method. First, we will substitute the functions to the formula and apply the integration by parts. Then, find the differentiation of the logarithmic expression by applying the chain rule of differentiation. Then, we will integrate the expression by applying the power rule of integration. Then, we will apply the exponent rule of logarithms to multiply the exponent to the logarithm expression.

Formula used:
Integration by parts is given by:
∫f(x)g(x)dx=f(x)∫g(x)dx−(∫f′(x)∫g(x)dx)dx\int {f\left( x \right)g\left( x \right)dx} = f\left( x \right)\int {g\left( x \right)dx} - \left( {\int {f'\left( x \right)} \int {g\left( x \right)dx} } \right)dx

The power rule of the integration is given by:

∫xndx=xn+1n+1\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}

The chain rule of differentiation is given by:

ddx(xnx)=nxn−1ddxnx\dfrac{d}{{dx}}\left( {{x^{nx}}} \right) = n{x^{n - 1}}\dfrac{d}{{dx}}nx

The differentiation of logarithmic function, ln⁡(x)\ln \left( x \right) is given by:

ddxln⁡(x)=1x\dfrac{d}{{dx}}\ln \left( x \right) = \dfrac{1}{x}

Complete step by step solution:
Let the integral be ∫xln⁡(x)2dx\int {x\ln {{\left( x \right)}^2}} dx
Here, f(x)=ln⁡(x)2f\left( x \right) = \ln {\left( x \right)^2} and g(x)=xg\left( x \right) = x. Apply the integration by parts method to find the integral.

⇒ln⁡(x)2∫xdx−(∫(ln⁡(x)2)′∫xdx)dx \Rightarrow \ln {\left( x \right)^2}\int {xdx} - \left( {\int {{{\left( {\ln {{\left( x \right)}^2}} \right)}^\prime }} \int {xdx} } \right)dx
Apply the chain rule of differentiation to the expression (ln⁡(x)2)′{\left( {\ln {{\left( x \right)}^2}} \right)^\prime }

⇒(ln⁡(x)2)′=1x2(2x)=2x \Rightarrow {\left( {\ln {{\left( x \right)}^2}} \right)^\prime } = \dfrac{1}{{{x^2}}}\left( {2x} \right) = \dfrac{2}{x}

⇒ln⁡(x)2∫xdx−(∫2x∫xdx)dx \Rightarrow \ln {\left( x \right)^2}\int {xdx} - \left( {\int {\dfrac{2}{x}} \int {xdx} } \right)dx

Apply the power rule of integration to the expression.

⇒ln⁡(x)2x22−2(∫1x×x22)dx \Rightarrow \ln {\left( x \right)^2}\dfrac{{{x^2}}}{2} - 2\left( {\int {\dfrac{1}{x}} \times \dfrac{{{x^2}}}{2}} \right)dx

Simplify the expression.

⇒ln⁡(x)2x22−2∫x2dx\Rightarrow \ln {\left( x \right)^2}\dfrac{{{x^2}}}{2} - 2\int {\dfrac{x}{2}dx}

⇒ln⁡(x)2x22−∫xdx\Rightarrow \ln {\left( x \right)^2}\dfrac{{{x^2}}}{2} - \int {xdx}

Apply the power rule of integration to the expression, ∫xndx=xn+1n+1\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}

⇒ln⁡(x)2x22−x22 \Rightarrow \ln {\left( x \right)^2}\dfrac{{{x^2}}}{2} - \dfrac{{{x^2}}}{2}

Rewrite the expression by applying exponent rule of logarithms to the expression ln⁡(x)2\ln {\left( x \right)^2}, ln⁡ab=bln⁡a\ln {a^b} = b\ln a

⇒ln⁡(x)2x22−x22=2ln⁡(x)×x22−x22 \Rightarrow \ln {\left( x \right)^2}\dfrac{{{x^2}}}{2} - \dfrac{{{x^2}}}{2} = 2\ln \left( x \right) \times \dfrac{{{x^2}}}{2} - \dfrac{{{x^2}}}{2}

Cancel out the common terms, we get:

⇒x2ln⁡(x)−x22+C \Rightarrow {x^2}\ln \left( x \right) - \dfrac{{{x^2}}}{2} + C
Final answer: Hence the value of the expression, ∫xln⁡(x)2dx\int {x\ln {{\left( x \right)}^2}} dx is x2ln⁡(x)−x22+C{x^2}\ln \left( x \right) - \dfrac{{{x^2}}}{2} + C

Note: Students please note that when the function is in the form of multiplication or division, then we will apply the integration by parts method. On applying the integration by parts, the values of f(x)f\left( x \right) and g(x)g\left( x \right) must be carefully chosen so that we find the function of LHS in the right hand side and we can combine them to simplify the expression.