## Calculus, 10th Edition (Anton)

$\frac{(4x-3)^{10}}{40}+C$
Use the substitution $u=4x-3$. This means $du=4dx$, so $dx=\frac{1}{4} du$. $\int (4x-3)^9dx$ $=\int u^9*\frac{1}{4} du$ $=\int \frac{1}{4}u^9 du$ $=\frac{1}{4}*\frac{u^{10}}{10}+C$ $=\frac{u^{10}}{40}+C$ $=\frac{(4x-3)^{10}}{40}+C$