Answer
$$\int r^4\Big(7-\frac{r^5}{10}\Big)^3dr=-\frac{1}{2}\Big(7-\frac{r^5}{10}\Big)^4+C$$
Work Step by Step
$$A=\int r^4\Big(7-\frac{r^5}{10}\Big)^3dr$$
We set $u=7-\frac{r^5}{10}$
Then $$du=-\frac{5r^4}{10}dr=-\frac{r^4}{2}dr$$
That means $$r^4dr=-2du$$
Therefore, $$A=-2\int u^3du=-\frac{2u^4}{4}+C=-\frac{u^4}{2}+C$$ $$A=-\frac{1}{2}\Big(7-\frac{r^5}{10}\Big)^4+C$$