## Calculus: Early Transcendentals 8th Edition

$$\int(x^2+1)(x^3+3x)^4dx=\frac{(x^3+3x)^5}{15}+C$$
$$A=\int(x^2+1)(x^3+3x)^4dx$$ Let $u=x^3+3x$. We would have $du=3x^2+3dx=3(x^2+1)dx$. Therefore, $(x^2+1)dx=\frac{1}{3}du$ Substitute into $A$, we have $$A=\frac{1}{3}\int u^4du$$ $$A=\frac{1}{3}\frac{u^5}{5}+C$$ $$A=\frac{u^5}{15}+C$$ $$A=\frac{(x^3+3x)^5}{15}+C$$