Question

What is the integral of $ \int{{{\ln }^{2}}\left( x \right)dx} $ ?

Answer 1

Hint: We first discuss the integration by parts method. Integration by parts method is usually used for the multiplication of the functions and their integration. We have to apply the theorem twice. We take $ u=1,v={{\ln }^{2}}x $ for integration $ \int{{{\ln }^{2}}\left( x \right)dx} $ . Then we take $ u=1,v=\ln x $ for integration $ \int{\ln \left( x \right)dx} $ . use the formulas $ \int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c $ , $ \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} $ .

Complete step-by-step answer:
We need to find the integration of $ \int{{{\ln }^{2}}\left( x \right)dx} $ using integration by parts method.
Let’s assume $ f\left( x \right)=g\left( x \right)h\left( x \right) $ .
We need to find the integration of
$ \int{f\left( x \right)dx}=\int{g\left( x \right)h\left( x \right)dx} $ .
We take $ u=g\left( x \right),v=h\left( x \right) $ . This gives $ \int{f\left( x \right)dx}=\int{uvdx} $ .
The theorem of integration by parts gives
$ \int{uvdx}=u\int{vdx}-\int{\left( \dfrac{du}{dx}\int{vdx} \right)dx} $ .
For our integration $ \int{{{\ln }^{2}}\left( x \right)dx} $ , we take $ u=1,v={{\ln }^{2}}x $ .
Now we complete the integration
\[\int{{{\ln }^{2}}\left( x \right)dx}={{\ln }^{2}}\left( x \right)\int{dx}-\int{\left( \dfrac{d\left( {{\ln }^{2}}\left( x \right) \right)}{dx}\int{dx} \right)dx}\].
We have the differentiation formula for
$ u={{\ln }^{2}}\left( x \right) $
where $ \dfrac{du}{dx}=\dfrac{d}{dx}\left( {{\ln }^{2}}\left( x \right) \right)=\dfrac{2\ln x}{x} $
The integration formula for
$ \int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c $ .
We apply these formulas to complete the integration and get
\[\int{{{\ln }^{2}}\left( x \right)dx}=x{{\ln }^{2}}\left( x \right)-\int{\left( \dfrac{2\ln x}{x}\times x \right)dx}=x{{\ln }^{2}}\left( x \right)-2\int{\ln xdx}\].
We have one more integration part.
To complete that we use again by parts $ \int{\ln xdx}=x\ln x-x $ .
So, the final form is
\[\int{{{\ln }^{2}}\left( x \right)dx}=x{{\ln }^{2}}\left( x \right)-2x\left( \ln x-1 \right)=x\left( {{\ln }^{2}}\left( x \right)-2\ln x+2 \right)\].
Here $ c $ is the integral constant.
Therefore, the integration by parts of $ \int{{{\ln }^{2}}\left( x \right)dx} $ gives \[x\left( {{\ln }^{2}}\left( x \right)-2\ln x+2 \right)\].
So, the correct answer is “\[x\left( {{\ln }^{2}}\left( x \right)-2\ln x+2 \right)\]+ C.”.

Note: In case one of two functions are missing and we need to form the by parts method, we will take the multiplying constant 1 as the second function. But we need to remember that we won’t perform by parts by taking $ u=1 $ .