int frac{d x}{2 e^{2 x}+3 e^x+1}= | vedclass

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(A) Let $I = \int \frac{dx}{2e^{2x} + 3e^x + 1}$.Put $e^x = t$,then $e^x dx = dt$,which implies $dx = \frac{dt}{t}$.Substituting these into the integr

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$x+\log \left(e^x+1\right)-2 \log \left(2 e^x+1\right)+c,$ where $c$ is the constant of integration

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(A) Let $I = \int \frac{dx}{2e^{2x} + 3e^x + 1}$.Put $e^x = t$,then $e^x dx = dt$,which implies $dx = \frac{dt}{t}$.Substituting these into the integral,we get $I = \int \frac{dt}{t(2t^2 + 3t + 1)} = \int \frac{dt}{t(2t+1)(t+1)}$.Using partial fractions: $\frac{1}{t(2t+1)(t+1)} = \frac{A}{t} + \frac{B}{2t+1} + \frac{C}{t+1}$.$1 = A(2t+1)(t+1) + Bt(t+1) + Ct(2t+1)$.For $t=0$,$1 = A(1)(1) \implies A=1$.For $t=-1$,$1 = C(-1)(-2+1) = C(-1)(-1) = C \implies C=1$.For $t=-1/2$,$1 = B(-1/2)(1/2) = -B/4 \implies B=-4$.Thus,$I = \int (\frac{1}{t} - \frac{4}{2t+1} + \frac{1}{t+1}) dt$.$I = \log|t| - 4 \cdot \frac{1}{2} \log|2t+1| + \log|t+1| + c$.$I = \log|e^x| - 2 \log|2e^x+1| + \log|e^x+1| + c$.$I = x + \log(e^x+1) - 2 \log(2e^x+1) + c$.

$\int \frac{d x}{2 e^{2 x}+3 e^x+1}=$

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