Evaluate the following integral. Integral from nothing to nothing StartFraction 3 e Superscript x Over e Superscript 2 x Baseline plus 2 e Superscript x Baseline plus 1 EndFraction dx∫ 3ex e2x+2ex+1dx Integral from nothing to nothing StartFraction 3 e Superscript x Over e Superscript 2 x Baseline plus 2 e Superscript x Baseline plus 1 EndFraction dx∫ 3ex e2x+2ex+1dxequals

Answer:[tex]\int \frac{3e^x}{e^{2x}+2e^x+1}dx=-\frac{3}{e^x+1}+C[/tex]Step-by-step explanation:To find this integral [tex]\int \frac{3e^x}{e^{2x}+2e^x+1}dx[/tex] you must:1. Take the constant out: [tex]3\cdot \int \frac{e^x}{e^{2x}+2e^x+1}dx[/tex]2. Factor [tex]{e^{2x}+2e^x+1}=(e^{x}+1)^2[/tex][tex]3\cdot \int \frac{e^x}{\left(e^x+1\right)^2}dx[/tex]3. Apply u-substitution [tex]u=e^x+1[/tex][tex]3\cdot \int \frac{e^x}{\left(u\right)^2}dx[/tex][tex]u=e^x+1\\du=e^xdx\\dx=\frac{du}{e^x}[/tex][tex]3\cdot \int \frac{e^x}{\left(u\right)^2}\frac{du}{e^x} \\\\3\cdot \int \frac{1}{u^2}du[/tex][tex]3\cdot \int \:u^{-2}du[/tex]4. Apply the Power Rule [tex]\int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1[/tex][tex]3\cdot \frac{u^{-2+1}}{-2+1}[/tex]5. Substitute back [tex]u=e^x+1[/tex][tex]3\cdot \frac{\left(e^x+1\right)^{-2+1}}{-2+1}=3\cdot -\left(e^x+1\right)^{-1}=-\frac{3}{e^x+1}[/tex]6. Add a constant to the solution[tex]-\frac{3}{e^x+1}+C[/tex]