Basic Integral

Let the given definite integral be $I$ . We'll first observe the behavior of the function $f(t)=1+\left \lfloor \log_{10}t \right \rfloor$ for $t\in [1,x]$ where $x$ is our required value.

We have, $f(t)=\begin{cases} 1~\forall t\in [1,10)\\ 2~\forall t\in [10,10^2)\\3~\forall t\in [10^2,10^3)\\ \vdots \\ (n+1)~\forall t\in [10^n,10^{n+1})~,n\in \mathbb{Z_0^+}\end{cases}$

Also, we can note that $x$ has to be a positive value $\gt 1$ because $f(x)$ is non-decreasing. Now,

$\large \displaystyle I=\int_1^{10} \dfrac{dt}{f(t)}+\int_{10}^{10^2} \dfrac{dt}{f(t)}+\cdots + \int_{10^p}^x \dfrac{dt}{f(t)}$

where $\exists p\in \mathbb{Z_0^+}\mid 10^p\leq x \leq 10^{p+1}$

The integral can be written using the behavior of $f(x)$ as:

$\large I=\displaystyle \int_1^{10}\,dt+\int_{10}^{10^2}\dfrac{1}{2}\,dt+\ldots+\int_{10^p}^x \dfrac{1}{p+1}\,dt \\ \large \implies I=[t]_1^{10}+\dfrac{1}{2}[t]_{10}^{10^2}+\ldots+\dfrac{1}{p+1}[t]_{10^p}^x$

We can observe that for $I=2005$ , we must have $p=3$ because $p=2$ would make $I=54\lt 2005$ and $p=4$ would make $I=2604\gt 2005$ . So, using $p=3$ , we can evaluate the integral as:

$I=(10-1)+\dfrac{1}{2}(100-10)+\dfrac{1}{3}(1000-100)+\dfrac{1}{4}(x-1000)=2015 \\ \implies 9+45+300+\dfrac{x-1000}{4}=2005 \implies \boxed{x=7604}$