Using u-substitution on an integral

If $u=\ln(x)+2$, then $x=e^{u-2}$ and $\mathrm dx=e^{u-2}\,\mathrm du$. The integral then becomes

$$\int_1^3 \frac{x^4}{\ln(x)+2}\,\mathrm dx=\int_2^{\ln(3)+2}\frac{e^{4(u-2)}}u(e^{u-2}\,\mathrm du)=\int_2^{\ln(3)+2}\frac{e^{5u-10}}u\,\mathrm du$$

The missing link between this and your solution is accounting for how $\mathrm dx$ is transformed.

If you take $u = \ln(x) +2$, then notice that $du = 1/x \cdot dx$. But, from the first, it follows that $x = e^{u-2}$. Thus, solving for $dx = x \cdot du$ and using that, we have

$$\int \frac{x^4}{\ln(x) + 2} dx = \int \frac{(e^{u-2})^4}{u} \cdot e^{u-2} du$$