# Evaluate the indefinite integral. \int \frac{e^{t}dt}{e^{2t}+2e^{t}+1}

Evaluate the indefinite integral.
$\int \frac{{e}^{t}dt}{{e}^{2t}+2{e}^{t}+1}$
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Shannon Hodgkinson

Step 1
Given: $I=\int \frac{{e}^{t}dt}{{e}^{2t}+2{e}^{t}+1}$...(1)
for evaluating given integral we substitute
${e}^{t}=x$...(2)
now differentiating equation (2) with respect to t
so,

${e}^{t}=\frac{dx}{dt}$
${e}^{t}dt=dx$
now replacing with x in equation (1)
Step 2
so,

$=\int \frac{dx}{{x}^{2}+2\left(x\right)\left(1\right)+{1}^{2}}$
$=\int \frac{dx}{{\left(x+1\right)}^{2}}$

$=\frac{{\left(x+1\right)}^{2+1}}{-2+1}+c$
$=\frac{{\left(x+1\right)}^{-1}}{-1}+c$
$=-\frac{1}{x+1}+c$...(3)
Step 3
now replacing x with et in equation (3)
$I=-\frac{1}{{e}^{t}+1}+c$
hence, given integral is equal to $-\frac{1}{{e}^{t}+1}+c$.

###### Not exactly what you’re looking for?
Andrew Reyes

We have:
$\int \frac{{e}^{t}}{{e}^{2t}+2{e}^{t}+1}dt$
$\int \frac{1}{{u}^{2}+2u+1}du$
$\int \frac{1}{{\left(u+1\right)}^{2}}du$
$\int \frac{1}{{v}^{2}}dv$
Evaluate the integral
$-\frac{1}{v}$
$-\frac{1}{u+1}$
Substitute back
$-\frac{1}{{e}^{t}+1}$
$-\frac{1}{{e}^{t}+1}+C$