Let $X$ be a random variable s.t. for $v, b > 0$ and $C \geq 1$:

$$ P(X \geq t) \leq C\exp\left(-\frac{t^2}{2(v^2 + bt)} \right) $$

I am trying to show that $\mathbb{E}X \leq 2v(\sqrt{\pi} + \sqrt{\log C}) + 4b(1 + \log C)$.

In terms of progress I have made so far, I set $X = \int_0^\infty P(X \geq t)dt$, and have also made cases for when $v^2 > bt$ and otherwise (i.e. simplifying the exponentiated value to be $-t^2/4v^2$ in one case and $-t^2/4bt$ in the other).

This obtainis all the non $C$ values in the final inequality; However, I can't figure out how to squash the $C$ into the $\log C$ and $\sqrt{\log C}$ terms.