I have accrossed this integral when I run some of my computation in Wolfram alpha with many values of $t$ , Really seems to conjecture that :
$$\int_{0}^{t}\operatorname{erf}(x+\sqrt{1-\log (x)} )dx \sim t$$ for every $t$ , The key I dea i have used is the series expansion of that function $x+\sqrt{1-\log (x)}$ at $x=e$ as shown in the below image , by substitution in the titled function that's make the integral is very complicated and as a reason we do not have a closed form for it this make obstacle to me to get what i have conjuctred , Now my justification if the conjecture is true is that :$\lim_{x\to \infty}\operatorname{erf}(x+\sqrt{1-\log (x)} ) =\lim_{x\to e}\operatorname{erf}(x+\sqrt{1-\log (x)} )=1 $ , But this is not enough and is not clear at all , Now my question how do i show the titled equality if it is true ?
Too long for a comment.
automaticallyGenerated gave the correct explanation. Just let me elaborate considering $$f(x)=\operatorname{erf}(x+\sqrt{1-\ln(x)})$$ for which $$f'(x)=\frac{2}{\sqrt{\pi }}e^{-\left(x+\sqrt{1-\log (x)}\right)^2} \left(1-\frac{1}{2 x \sqrt{1-\log(x)}}\right)$$ So, the derivative cancels when $$2 x \sqrt{1-\log(x)}=1$$ that is to say when $$1+\frac 12 \log\left(\frac 1 {x^2} \right)=\frac 1 {4x^2}$$ the solution of which being $$x_*=\sqrt{2 W\left(\frac{e^2}{2}\right)}\approx 1.52260$$ where $W(.)$ is Lambert function. $$f(x_*)=0.998762$$
Edit
Interesting would be the Taylor expansion around $x=1$; this would be $$f(x)=\text{erf}(2)+\frac{x-1}{e^4 \sqrt{\pi }}-\frac{3 (x-1)^2}{4 e^4 \sqrt{\pi }}-\frac{(x-1)^3}{8 e^4 \sqrt{\pi }}+\frac{137 (x-1)^4}{192 e^4 \sqrt{\pi }}+O\left((x-1)^5\right)$$ making $$\int_0^t f(x)\,dx\approx\left(\text{erf}(2)-\frac{175}{192 e^4 \sqrt{\pi }}\right) t-\frac{35 t^2}{96 e^4 \sqrt{\pi }}+\frac{125 t^3}{96 e^4 \sqrt{\pi }}-\frac{143 t^4}{192 e^4 \sqrt{\pi }}+\frac{137 t^5}{960 e^4 \sqrt{\pi }}$$ which, numerically is $$0.985904 t-0.00376742 t^2+0.0134551 t^3-0.0076963 t^4+0.00147468 t^5$$