How do you evaluate the indefinite integral \[\int{\left( {{x}^{3}}+3x+1 \right)dx}\]?

Hint: Assume the value of the given integral as ‘I’. Break the integral into three parts and integral each of the terms: - \[{{x}^{3}},3x\] and 1. Use the basic integral formula, \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\] for \[n\ne -1\]. To use this formula for the constant term 1, write it as \[{{x}^{0}}\] and then evaluate. Add the constant of indefinite integration ‘C’ at last to get the answer.

Complete step by step answer:
Here, we have been provided with the function \[\left( {{x}^{3}}+3x+1 \right)\] and we are asked to find its integral. Let us assume its integral as I, so we have,
\[\Rightarrow I=\int{\left( {{x}^{3}}+3x+1 \right)dx}\]
Breaking the integral into three parts, one for each term, we have,
\[\Rightarrow I=\int{{{x}^{3}}dx}+\int{3xdx}+\int{1dx}\]
Now, we can write the constant term 1 as \[{{x}^{0}}\], so we have,
\[\Rightarrow I=\int{{{x}^{3}}dx}+\int{3{{x}^{1}}dx}+\int{{{x}^{0}}dx}\]
Now, applying the basic formula of integral given as: - \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\], we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{{{x}^{3+1}}}{3+1}+3\left( \dfrac{{{x}^{1+1}}}{1+1} \right)+\dfrac{{{x}^{0+1}}}{0+1} \\
 & \Rightarrow I=\dfrac{{{x}^{4}}}{4}+\dfrac{3{{x}^{2}}}{2}+x \\
\end{align}\]
Taking L.C.M., i.e., 4 and simplifying we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{{{x}^{4}}+6{{x}^{2}}+4x}{4} \\
 & \Rightarrow I=\dfrac{1}{4}\left( {{x}^{4}}+6{{x}^{2}}+4x \right) \\
\end{align}\]
Now, since the given integral was an indefinite integral and therefore we need to add a constant of integration (C) in the expression obtained for I. So, we get,
\[\Rightarrow I=\dfrac{1}{4}\left( {{x}^{4}}+6{{x}^{2}}+4x \right)+C\]
Hence, the above relation is our answer.

Note: One may note that the basic formula that we have used to find the integral I given as: - \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\] is invalid for n = -1. This is because in this case (n + 1) will become 0 and the integral will become undefined. So, when n = -1 then the function becomes \[\int{\dfrac{1}{x}dx}\] whose solution is \[\ln x\]. You must remember all the basic formulas of indefinite integrals and that for the different functions. At last, do not forget to add the constant of integration (C) as we are finding indefinite integral and not definite integral.