Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 329: 3

$\frac{(x^2+5)^{-3}}{-3}+c$

$u=x^2+5$, therefore $du=2xdx$. Let's rewrite the integral! $\int (2x(x^2+5)^{-4})dx=\int ((x^2+5)^{-4})2xdx$ We do the substitution! $\int ((x^2+5)^{-4})2xdx=\int u^{-4}du=\frac{u^{-3}}{-3}+c$ We re-do the substitution! $\frac{u^{-3}}{-3}+c=\frac{(x^2+5)^{-3}}{-3}+c$