Answer:
Step-by-step explanation:
Given;
[tex]f(x)=\frac{e^{2x}}{e^{2x}+3e^x+2}[/tex]
a)
substitute [tex]u=e^x\\du=e^x dx\\\\\int\frac{e^{2x}}{e^{2x}+3e^x+2}dx=\int\frac{e^x\dot e^x}{e^x^{2x}+3e^x+2}dx\\\\=\int\frac{udu}{u^2+3u+2}[/tex]
b)
Apply partial fraction in (a), we get;
[tex]\frac{u}{u^2+3u+2}=\frac{2}{u+2}-\frac{1}{u+1}\\\\\therefore u^2+3u+2\\=u^2+2u+u+2\\=u(u+2)+1(u+1)\\=(u+2)(u+1)\\\\Now\,\int\frac{u}{u^2+3u+2}\,du=\int\frac{2}{u+2}du-\int\frac{1}{u+1}du\\\\=2ln|u+2|-ln|u+1|+c\\=2ln|e^x+2|-lm|e^x+1|+c[/tex]
where C is an arbitrary constant