## Calculus: Early Transcendentals 8th Edition

$$A=\frac{x^{2.3}}{2.3}+2x^{3.5}+C$$
$$A=\int(x^{1.3}+7x^{2.5})dx$$ $$A =\int{(x^{1.3})dx}+7\int(x^{2.5})dx$$ According to Table 1, we know that $\int(x^n)dx=\frac{x^{n+1}}{n+1}+C$ $(n\ne-1)$. Therefore, $$A=\frac{x^{2.3}}{2.3}+\frac{7x^{3.5}}{3.5}+C$$ $$A=\frac{x^{2.3}}{2.3}+2x^{3.5}+C$$