$$f(x) = \begin{cases}\frac{\alpha}{\beta^\alpha}x^{\alpha-1}e^{-(x/\beta)^{\alpha}} & x \ge0\\ 0 & x<0\end{cases}$$
Also $\alpha, \beta>0$.
I tried substituting $u=-(x/\beta)^{\alpha}$ and then solving $\int_0^{x}\frac{\alpha}{\beta^\alpha}x^{\alpha-1}e^{-(x/\beta)^{\alpha}}dx$ but got total mess solving the integral
Not sure why you got a total mess. Maybe part of the problem is you've confused dummy variables with variables in the limits. We want $$ \int_0^x \frac{\alpha}{\beta^\alpha}t^{\alpha-1}e^{-(t/\beta)^\alpha}\,dt.$$ Subbing in $u = (t/\beta)^\alpha,$ gives $du=\frac{\alpha}{\beta^\alpha}t^{\alpha-1}dt$ so the substitution reduces the integral to $$ \int_0^{(x/\beta)^\alpha}e^{-u}\,du,$$ which is doable.