Definite integration: $int_1^2 frac{(1+ln x)^5}{x}dx$

Evaluate following definite integral:
$$int_1^2 frac{(1+ln x)^5}{x}dx$$
I solved this integral with a substitution rule: $u=ln x$, $du=dx/x$

$$int_1^2 frac{(1+ln x)^5}{x}dx=int_0^{ln 2} (1+u)^5 du$$

solved and got an answer $1.97$. But I got a different result from that others i.e; $3.75$.

I am just wondering am I right or wrong?

Thanks in advance.