Explain how you will integrate a given function:
 $ \int {(x + 3)dx} $

Hint: Observe the question carefully. In case an integral is given, we must first understand what all kinds of terms are given. Then we must remember the basic rules and formulas we use for integration of constants and variables. Always integrate the given function with respect to the variable that is given in the question, here it is with respect to $ dx $ . So every term with $ x $ will be integrated with respect to $ dx $ and the other terms will be treated as constants.

Complete step-by-step answer:
First we must always note down what is given to us and what we need to do.
The function given here is an integral function: $ \int {(x + 3)dx} $
We are asked to integrate it.
Now consider the given integral $ \int {(x + 3)dx} $ .
Each term has to be treated separately and each term will be integrated with respect to $ dx $ .
The formula we must use for integration of any terms is given below:
 $ \Rightarrow \int {(a{x^n})dx = \dfrac{a}{{n + 1}}{x^{n + 1}}} + c $ ; here $ a = $ given constant within the integral, $ c = $ integral constant which must be put after integration of any function.
Now look at each term in the integral, since it is an additional operation, the terms can be treated separately.
For the first term that is $ x $ the integral will be:
 $ \int {xdx} = \dfrac{{{x^2}}}{2} + c_1 $
Then look at the second term, that is: $ 3 $ , so integral of that term;
 $ \int {3{x^0}dx = 3x + c_2} $
Now combining the two equations we get;
 $ \Rightarrow \int {(x + 3)dx = \dfrac{{{x^2}}}{2} + 3x + c_1} + c_2 $
But constants can be combined to become a single term so the final integral will look like:
 $ \therefore \int {(x + 3)dx = \dfrac{{{x^2}}}{2} + 3x + k} $
So, the correct answer is “ $ \therefore \int {(x + 3)dx = \dfrac{{{x^2}}}{2} + 3x + k} $ ”.

Note: We must remember that integration is a very important topic, with very many applications. The different sections where it is used are in geometric applications, then for physical applications. In specific they are used to calculate area and volumes of different figures, then also for the decay and growth models. In Physics we use it to calculate the distance, acceleration and velocity also.