How do you integrate $\ln ({x^2}\,\, + \,\,\,4)$ ?

Hint: To solve this question, we must first have knowledge of the basics of integration and the techniques that we use to solve the integrals. And we can solve this question in a very simple manner by using the Integration by parts and substitution method and then only we can conclude the correct answer.

Complete Step by Step answer:
Before we move forward with the solution of this given question, let us first understand some basic concepts:
As we know that in Integral calculus, integration by substitution method which is also known as u-substitution method or change of variables method. And it is a method for evaluating integrals and antiderivatives.

In the Substitution method we basically substitute the variables part (which is heavy or complex to solve directly using the integration formula) with a temporary variable or in terms of a temporary variable which will eventually remove the complexities (usually we use the variable ‘u’ or ‘t’) and then convert the integral in the form of that variable and simply solve for that variable and when we get the answer in terms of that new variable, then we again substitute the values of new assumed variable with the original one. And hence we get our required answer.

The formula of integration by parts is:
$\smallint {1^{st}}fun \times {2^{nd}}fun\,\, = \,\,{2^{nd}}fun \times \smallint {1^{st}}fun\,\, - \,\,\smallint \left( {\smallint {1^{st}}fun \times d\left( {{2^{nd}}fun} \right)} \right)$
Where, fun = function and d = differentiation.
Step 1: We will use integration by parts:
Let $I\,\, = \,\,\smallint \ln ({x^2}\,\, + \,\,\,4) \times 1\,\,dx$
Here we assume that out first function is 1 and our second function is $\ln ({x^2}\,\, + \,\,\,4)$
By applying integration by parts on $I\,\, = \,\,\smallint \ln ({x^2}\,\, + \,\,\,4) \times 1\,\,dx$ we get:
$ \Rightarrow I\,\, = \,\,x\ln ({x^2}\,\, + \,\,\,4)\,\, - \,\,\smallint x \times \dfrac{{2x}}{{{x^2}\,\, + \,\,4}}dx$
And in the above step we have calculated the differentiation of second function we have used substitution:
$
\Rightarrow u\,\, = \,\,\ln ({x^2}\,\, + \,\,4) \\
  then,\,\,du\, = \,\dfrac{1}{{{x^2}\,\, + \,\,4}} \times 2x \times dx\,\, = \,\,\dfrac{{2x}}{{{x^2}\,\, + \,\,4}}dx \\
$

Step 2:
$ \Rightarrow I\,\, = \,\,x\ln ({x^2}\,\, + \,\,\,4)\,\, - \,\,2\smallint \dfrac{{{x^2}}}{{{x^2}\,\, + \,\,4}}dx$
$ \Rightarrow I\,\, = \,\,x\ln ({x^2}\,\, + \,\,\,4)\,\, - \,\,2\smallint \dfrac{{{x^2} + 4 - 4}}{{{x^2}\,\, + \,\,4}}dx \\
\Rightarrow I\,\, = \,\,x\ln ({x^2}\,\, + \,\,\,4)\,\, - \,\,2\smallint \left( {1 - \dfrac{4}{{{x^2}\,\, + \,\,4}}} \right)dx \\
$
$
\Rightarrow I\,\, = \,\,x\ln ({x^2}\,\, + \,\,\,4)\,\, - \,\,2\smallint \dfrac{{{x^2} + 4 - 4}}{{{x^2}\,\, + \,\,4}}dx \\
\Rightarrow I\,\, = \,\,x\ln ({x^2}\,\, + \,\,\,4)\,\, - \,\,2\left( {x - 4 \times \dfrac{1}{2}{{\tan }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)dx\,\, + \,\,C \\
$

And hence the required answer is: $I\,\, = \,\,x\ln ({x^2}\,\, + \,\,\,4)\,\, - \,\,2x + 4{\tan ^{ - 1}}\left( {\dfrac{x}{2}} \right)dx\,\, + \,\,C$

Note: 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.