Irvin Dukes

2021-12-20

Evaluate the following integrals.
$\int \frac{{e}^{x}}{{e}^{2x}+2{e}^{x}+1}dx$

kalupunangh

Expert

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

Barbara Meeker

Expert

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

nick1337

Expert

$\int \frac{{e}^{x}}{{e}^{2x}+2{e}^{x}+1}dx$
Let us put the expression exp (x) under the sign of the differential, ie:
${e}^{x}\ast dx=d\left({e}^{x}\right),t={e}^{x}$
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
$\frac{1}{\left(x+1{\right)}^{2}}=\frac{1}{\left(x+1{\right)}^{2}}$
use the method of decomposition into simplest elements. Let us expand the function into the simplest terms:
$\frac{1}{\left(x+1{\right)}^{2}}=\frac{A}{x+1}+\frac{B}{\left(x+1{\right)}^{2}}=\frac{A\left(x+1\right)+B}{\left(x+1{\right)}^{2}}$
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;
$\frac{1}{\left(x+1{\right)}^{2}}=\frac{0}{x+1}+\frac{1}{\left(x+1{\right)}^{2}}$
We calculate the table integral:
$\int \frac{1}{\left(x+1{\right)}^{2}}\ast dx=-\frac{1}{x+1}$
NSK
$-\frac{1}{x+1}+C$
To write down the final answer, it remains to substitute exp (x) instead of t.
$-\frac{1}{{e}^{x}+1}+C$

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