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

Step 1
Given: ${\displaystyle I=\int \frac{e^{t}dt}{e^{2t}+2e^t+1}}$...(1)
for evaluating given integral we substitute
${\displaystyle e^t=x}$...(2)
now differentiating equation (2) with respect to t
so,
${\displaystyle \frac{d}{dt}\left(e^t\right)=\frac{d}{dt}\left(x\right)(\because \frac{d}{dx}\left(e^x\right)=e^x)}$
${\displaystyle e^t=\frac{dx}{dt}}$
${\displaystyle e^{t}dt=dx}$
now replacing ${\displaystyle e^{t}dtaskwithdx\text{and}{e}^t}$ with x in equation (1)
Step 2
so,
${\displaystyle I=\int \frac{dx}{x^2+2x+1}(\because a^2+2ab+b^2={(a+b)}^2)}$
${\displaystyle =\int \frac{dx}{x^2+2\left(x\right)\left(1\right)+1^2}}$
${\displaystyle =\int \frac{dx}{{(x+1)}^2}}$
${\displaystyle =\int {(x+1)}^{-2}dx(\because \int {(x+a)}^{n}dx=\frac{{(x+a)}^{n+1}}{n+1}+c)}$
${\displaystyle =\frac{{(x+1)}^{2+1}}{-2+1}+c}$
${\displaystyle =\frac{{(x+1)}^{-1}}{-1}+c}$
${\displaystyle =-\frac{1}{x+1}+c}$...(3)
Step 3
now replacing x with et in equation (3)
${\displaystyle I=-\frac{1}{e^t+1}+c}$
hence, given integral is equal to ${\displaystyle -\frac{1}{e^t+1}+c}$.