Answer
$$\int(1-2x)^9dx=-\frac{(1-2x)^{10}}{20}+C$$
Work Step by Step
$$A=\int(1-2x)^9dx$$
Let $u=1-2x$
Then $du=(1-2x)'dx=-2dx$. So $dx=-\frac{1}{2}du$
Substitute into $A$, we have $$A=\int u^9(-\frac{1}{2})du$$ $$A=-\frac{1}{2}\int u^9du$$ $$A=-\frac{1}{2}\frac{u^{10}}{10}+C$$ $$A=-\frac{u^{10}}{20}+C$$ $$A=-\frac{(1-2x)^{10}}{20}+C$$
