Evaluate the following integrals. ∫exe2x+2ex+1dx

Step 1
We have to evaluate the integral:
${\displaystyle \int \frac{e^x}{e^{2x}+2e^x+1}dx}$
Rewriting the integral,
${\displaystyle \int \frac{e^x}{e^{2x}+2e^x+1}dx=\int \frac{e^x}{\left(e^x\right)+2e^x+1}dx}$
This will be solved by substitution method,
so assuming ${\displaystyle t=e^x}$
differentiating,
${\displaystyle \frac{dt}{dx}=\frac{d{e}^x}{dx}}$
${\displaystyle \frac{dt}{dx}=e^x}$
${\displaystyle dt=e^{x}dx}$
Step 2
Substituting above values in the integral,
${\displaystyle \int \frac{e^x}{{\left(e^x\right)}^2+2e^x+1}dx=\int \frac{dt}{t^2+2t+1}}$
${\displaystyle =\int \frac{dt}{{\left(t\right)}^2+2\left(t\right)\left(1\right)+{\left(1\right)}^2}}$
${\displaystyle =\int \frac{dt}{{(t+1)}^2}}$
${\displaystyle =-\frac{1}{t+1}+C}$ (since ${\displaystyle \int \frac{1}{{(x+a)}^2}=-\frac{1}{x+a}+C}$)
Where, C is an arbitrary constant.
Hence, value of integral is ${\displaystyle -\frac{1}{e^x+1}+C}$.

It is required to calculate:
${\displaystyle \int \frac{e^x}{e^{2x}+2e^x+1}dx}$
Substitution ${\displaystyle u=e^x\Rightarrow \frac{du}{dx}=e^x\Rightarrow dx=e^{-x}du}$, we use:
${\displaystyle e^{2x}=u^2}$
${\displaystyle =\int \frac{1}{u^2+2u+1}du}$
Let us factorize:
${\displaystyle =\int \frac{1}{{(u+1)}^2}du}$
Substitution ${\displaystyle v=u+1\Rightarrow \frac{dv}{du}=1\Rightarrow du=dv:}$
${\displaystyle =\int \frac{1}{v^2}dv}$
Integral of a power function:
${\displaystyle \int v^{n}dv=\frac{v^{n+1}}{n+1}}$ at n=-2:
${\displaystyle =-\frac{1}{v}}$
Reverse replacement v=u+1:
${\displaystyle =-\frac{1}{u+1}}$
Reverse replacement ${\displaystyle u=e^x:}$
${\displaystyle =-\frac{1}{e^x+1}}$
Problem solved:
${\displaystyle \int \frac{e^x}{e^{2x}+2e^x+1}dx}$
${\displaystyle =-\frac{1}{e^x+1}+C}$

$\int \frac{e^x}{e^{2x}+2e^x+1}dx$
Let us put the expression exp (x) under the sign of the differential, ie:
$e^x\ast dx=d(e^x),t=e^x$
Then the original integral can be written as follows:
$\int \frac{1}{t^2+2\ast t+1}\ast dt$
$\int \frac{1}{x^2+2\ast x+1}\ast dx$
Let's
$\frac{1}{(x+1{)}^2}=\frac{1}{(x+1{)}^2}$
use the method of decomposition into simplest elements. Let us expand the function into the simplest terms:
$\frac{1}{(x+1{)}^2}=\frac{A}{x+1}+\frac{B}{(x+1{)}^2}=\frac{A(x+1)+B}{(x+1{)}^2}$
Equate the numerators and take into account that the coefficients at the same powers of x , standing on the left and on the right, must coincide:
1=A(x+1)+B
x: A=0
1: A+B=1
Solving it, we find:
A=0; B=1;
$\frac{1}{(x+1{)}^2}=\frac{0}{x+1}+\frac{1}{(x+1{)}^2}$
We calculate the table integral:
$\int \frac{1}{(x+1{)}^2}\ast dx=-\frac{1}{x+1}$
NSK
Answer:
$-\frac{1}{x+1}+C$
To write down the final answer, it remains to substitute exp (x) instead of t.
$-\frac{1}{e^x+1}+C$