## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{x^4}{4}+\dfrac{5x^2}{2}-7x+c$
Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ where $c$ is a constant of proportionality. Now, $\int x^3+5x-7 dx=\dfrac{x^{3+1}}{3+1}+5 \dfrac{x^{1+1}}{1+1}-7x+c$ or, $=\dfrac{x^4}{4}+\dfrac{5x^2}{2}-7x+c$