Answer
$\dfrac{x^4}{4}+\dfrac{5x^2}{2}-7x+c$
Work Step by Step
Use formula $\int x^{a} dx=\dfrac{x^{(a+1)}}{(a+1)}+c$
where $c$ refers to a constant of proportionality.
Then, we have $\int x^3+5x-7 dx=\dfrac{x^{3+1}}{3+1}+(5) \dfrac{x^{1+1}}{(1+1)}-7x+c \\ \implies \dfrac{x^{4}}{4}+(5) \dfrac{x^{2}}{2}-7x+c =\dfrac{x^4}{4}+\dfrac{5x^2}{2}-7x+c$
