An elementary application of Integral identity

Given that $$\mathbb{P}\bigg(|X|>C\cdot (\sigma\sqrt{\log{n}+u}+K(\log{n}+u))\bigg)<2e^{-u},$$ show that $$\mathbb{E}\bigg(|X|\bigg) where $$C,C'>0$$ are some absolute constant, and $$K, \sigma>0$$.

The textbook (Vershynin high dimensional probability exercise 5.4.11) says that I'm supposed to use integral identity $$\mathbb{E}\bigg(|X|\bigg)=\int_0^\infty \mathbb{P}(|X|>a) da$$ to show this. What I did is to do a change of variable $$a=C(\sigma\sqrt{\log{n}+u}+K(\log{n}+u))$$. Then I have $$\mathbb{E}\bigg(|X|\bigg)\le \int_{-\log{n}}^\infty 2e^{-u} C(K+\sigma\frac{1}{\sqrt{\log n +u}})du=nCK+nC\sigma\int_0^\infty e^{-t^2}dr$$ which does not solve the problem as it is linear to $$n$$ instead of $$\log n$$ as required.

I think there are two ways to do this. First is to use the $$\max(1,U(t))$$ integral identity trick: $$U(t)$$ is the tail bound and we solve $$U(t)=1$$ to find out where we should $$\int_0^r1dt+\int_r^\infty U(t)dt$$.
But specific to this question of mine, it is easiest to consider u as a random variable U such that $$|X|=R(U)$$ ($$R(U)$$ is RHS of the tail bound). This way, since R(U) is monotone increasing, $$P(U>u)=P(|X|>R(u))\le 2e^{-t^2/2}.$$ Then U is just subgaussian with constant subgaussian norm. Take expectation on both sides (and use Jensen's inequality to put expectation inside sqrt).