If $\int\limits_{2 x^2}^{x^3}(\ln x) f(t) d t=x^2-2 x+5$, then $f(8)=$

$\frac{1}{2 \ln 2}$

We have,

$\int\limits_{2 x^2}^{x^3}(\ln x) f(t) d t=x^2-2 x+5$

$\Rightarrow \frac{d}{d x}\left\{\int\limits_{2 x^2}^{x^3}(\ln x) f(t) d t\right\}=\frac{d}{d x}\left(x^2-2 x+5\right)$

$\Rightarrow \int\limits_{2 x^2}^{x^3} \frac{1}{x} f(t) d t+3 x^2(\ln x) f\left(x^3\right)-4 x(\ln x) f\left(2 x^2\right)=2 x-2$

Replacing $x=2$, we obtain

$12(\ln 2) f(8)-8(\ln 2) f(8)=2$

$\Rightarrow f(8)=\frac{1}{2 \ln 2}$