# How do I find the antiderivative of e^(2x) + 1?

Jan 27, 2015

I would use the idea of integral (indefinite) and the techniques connected with this procedure:
1) I can write:
$\int {e}^{2 x} + 1 \mathrm{dx}$
2) I can use the fact the the integral of a sum is equal to the sum of the integrals, giving:
$\int {e}^{2 x} \mathrm{dx} + \int 1 \mathrm{dx}$
3) I can use the fact that the integral of the exponential is equal to itself (but here we have to consider the exponent $2 x$ as well) and that $1$ can be written as ${x}^{0}$;
$\int {e}^{2 x} \mathrm{dx} + \int {x}^{0} \mathrm{dx} =$
${e}^{2 x} / 2 + x + c$

I also evaluate the integral of ${x}^{0}$ using the fact that the integral of ${x}^{n}$ is ${x}^{n + 1} / \left(n + 1\right)$

4) You can now check the result (the anti-derivative) obtained above deriving it to see if it gives the initial function ${e}^{2 x} + 1$.