Find the integral of $\frac{k+1}{4x^{2}+8}$ using the table of common integrals rule $\int a\mathrm{d}u=au$.
$$\frac{k+1}{4x^{2}+8}u$$
Simplify.
$$\frac{\left(k+1\right)u}{4x^{2}+8}$$
If $F\left(u\right)$ is an antiderivative of $f\left(u\right)$, then the set of all antiderivatives of $f\left(u\right)$ is given by $F\left(u\right)+C$. Therefore, add the constant of integration $C\in \mathrm{R}$ to the result.
