## Calculus, 10th Edition (Anton)

$=\frac{(a+bx)^{n+1}}{b(n+1)}+C$
Perform a u-substitution with $u=a+bx$. $du=b dx$, so $\frac{dx}{b}=dx$. This leaves $\frac{1}{b}\int{u^n}du$, which is $\frac{u^{n+1}}{b(n+1)}+C$. Converting back into terms of $x$ yields $=\frac{(a+bx)^{n+1}}{b(n+1)}+C$.