$$\int_0^{t_0}\frac{518.4935}{{66.9}-\frac{3.4A^2e^{at-at_0}}{0.029Ae^{at_0}+0.008A^3e^{3at_0}-0.726}}\mathrm{d}t$$ I calculated it into the following formula, but I don't know whether it is right or not, and I don't know how to calculate it $$\int_0^{t_0}\frac{15.0363115Ae^{2at_0}+4.147948A^3e^{4at_0}-376.426281e^{at_0}}{1.9401Ae^{2at_0}+0.5352A^3e^{4at_0}-48.5694e^{at_0}-3.4A^2e^{at}}\mathrm{d}t$$ This problem is so complicated that I can't calculate it at all. I need help. Thank you very much
Through your prompt, I made some new attempts. $$\int_0^{t_0}\frac{M}{{N}-\frac{PA^2}{UAe^{at_0}+VA^3e^{3at_0}-W}e^{at-at_0}}\mathrm{d}t$$ Let $$e^{at}=x$$ $$=\int_1^{e^{at_0}}\frac{M}{{ax}({N}-\frac{PA^2}{UAe^{2at_0}+VA^3e^{4at_0}-We^{at_0}}x)}\mathrm{d}x$$ $$=\frac{M}{Na}\int_1^{e^{at_0}}\frac{1}{x}+\frac{\frac{PA^2}{UAe^{2at_0}+VA^3e^{4at_0}-We^{at_0}}}{{N}-\frac{PA^2}{UAe^{2at_0}+VA^3e^{4at_0}-We^{at_0}}x}\mathrm{d}x$$ Is this correct
