# \int{x}^{3}+{3}\frac{{x}^{2}}{{x}^{2}}+{1}

$\int {x}^{3}+3\frac{{x}^{2}}{{x}^{2}}+1$

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Take the integral:
$\int \left({x}^{3}+4\right)dx$
Integrate the sum term by term and factor out constants:
= $\int {x}^{3}dx+4\int 1dx$
The integral of ${x}^{3}$ is $\frac{{x}^{4}}{4}:$
$=\frac{{x}^{4}}{4}+4\int 1dx$
The integral of 1 is x:
$=\frac{{x}^{4}}{4}+4x+\text{constant}$