## Calculus (3rd Edition)

$\frac{1}{4}(x-7)^{4}+C$
Substituting u=x-7 so that du=dx, we get $\int (x-7)^{3}dx= \int u^{3}du= \frac{u^{4}}{4}+C$ Undoing substitution, we have $\int (x-7)^{3} dx= \frac{1}{4}(x-7)^{4}+C$