How do you integrate \[\int{\left( {{x}^{2}}+1 \right)\log xdx}\] ?

Hint: In the given question, we have been asked to integrate the given expression. In order to integrate the given expression, we will be using the method of by-parts. The formula for integrating the given function by the method of by-parts is \[\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( f'\left( x \right)\int{g\left( x \right)} \right)dx}}\] .Then we will take \[\log x\] as the first function and \[\left( {{x}^{2}}+1 \right)\] as the second function. Now according to the rule we will solve the integration further by applying or using the rules of integration.

Formula used:
Integration by using the method of by-parts is as follows; \[\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( f'\left( x \right)\int{g\left( x \right)} \right)dx}}\] .
 \[\dfrac{d}{dx}\log x=\dfrac{1}{x}\]
Power rule of integration;
 \[\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}\]

Complete step by step solution:
We have given that,
 \[\int{\left( {{x}^{2}}+1 \right)\log xdx}\]
Let I be the integral,
 \[\Rightarrow I=\int{\left( {{x}^{2}}+1 \right)\log xdx}\]
Applying the formula of by parts i.e. \[\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( f'\left( x \right)\int{g\left( x \right)} \right)dx}}\]
Now,
Putting \[f\left( x \right)=\log x\] and \[g\left( x \right)={{x}^{2}}+1\]
Therefore,
 \[\Rightarrow I=\int{\left( {{x}^{2}}+1 \right)\log xdx}=\log x\int{\left( {{x}^{2}}+1 \right)dx}-\int{\left( \dfrac{d}{dx}\log x\int{\left( {{x}^{2}}+1 \right)dx} \right)dx}\]
Thus,
 \[\Rightarrow I=\log x\int{\left( {{x}^{2}}+1 \right)dx}-\int{\left( \dfrac{d}{dx}\log x\int{\left( {{x}^{2}}+1 \right)dx} \right)dx}\]
Applying the rules of integration, we will get
We know that the derivation of \[\dfrac{d}{dx}\log x=\dfrac{1}{x}\]
And using the integration rule i.e. \[\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}\]
 \[\Rightarrow I=\log x\left( \dfrac{{{x}^{3}}}{3}+x \right)-\int{\dfrac{1}{x}\left( \dfrac{{{x}^{3}}}{3}+x \right)}dx\]
Now, solving the above
 \[\Rightarrow I=\left( \dfrac{{{x}^{3}}}{3}+x \right)\log x-\int{\left( \dfrac{{{x}^{2}}}{3}+1 \right)}dx\]
Applying the rules of integration, we get
 \[\Rightarrow I=\left( \dfrac{{{x}^{3}}}{3}+x \right)\log x-\left( \dfrac{{{x}^{3}}}{3\times 3}+x \right)+C\]
Simplifying the above, we will get
 \[\Rightarrow I=\left( \dfrac{{{x}^{3}}}{3}+x \right)\log x-\dfrac{{{x}^{3}}}{9}-x+C\]
Therefore,
 \[\Rightarrow \int{\left( {{x}^{2}}+1 \right)\log xdx}=\left( \dfrac{{{x}^{3}}}{3}+x \right)\log x-\dfrac{{{x}^{3}}}{9}-x+C\] .
Hence, this is the required integration.
So, the correct answer is “ \[\Rightarrow \int{\left( {{x}^{2}}+1 \right)\log xdx}=\left( \dfrac{{{x}^{3}}}{3}+x \right)\log x-\dfrac{{{x}^{3}}}{9}-x+C\] ”.

Note: While solving these types of question, students always need to remember the formula for integration using the by-parts method i.e. \[\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( f'\left( x \right)\int{g\left( x \right)} \right)dx}}\] . Students mostly make mistakes while choosing the f(x) and g(x), so you should be very careful while choosing the f(x) and g(x) as it will make the given integration easy to solve easily. We should be well aware about the by-parts method of integration and we should also be well known about the application of integration. Students should be very careful while doing the calculation part of the given integral.