## Calculus: Early Transcendentals 8th Edition

$$\int x(2x+5)^8dx=\frac{1}{40}(2x+5)^{10}-\frac{5}{36}(2x+5)^9+C$$
$$A=\int x(2x+5)^8dx$$ We would try to break down the formula, but right now $(2x+5)^8$ is too hard to break down. We can apply substitution here. Let $u=2x+5$. Then $du=(2x+5)'dx=2dx$. Therefore, $dx=\frac{1}{2}du$ And since $u=2x+5$, we have $x=\frac{1}{2}(u-5)$ Substitute into $A$: $$A=\int\frac{1}{2}(u-5)u^8\frac{1}{2}du$$ $$A=\frac{1}{4}\int(u-5)u^8du$$ $$A=\frac{1}{4}\int(u^9-5u^8)du$$ $$A=\frac{1}{4}(\frac{u^{10}}{10}-\frac{5u^9}{9})+C$$ $$A=\frac{u^{10}}{40}-\frac{5u^9}{36}+C$$ $$A=\frac{1}{40}(2x+5)^{10}-\frac{5}{36}(2x+5)^9+C$$