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

Evaluate the following integrals.
$$\displaystyle\int{\frac{{{e}^{{{x}}}}}{{{e}^{{{2}{x}}}+{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}$$

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kalupunangh
Step 1
We have to evaluate the integral:
$$\displaystyle\int{\frac{{{e}^{{{x}}}}}{{{e}^{{{2}{x}}}+{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}$$
Rewriting the integral,
$$\displaystyle\int{\frac{{{e}^{{{x}}}}}{{{e}^{{{2}{x}}}+{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}=\int{\frac{{{e}^{{{x}}}}}{{{\left({e}^{{{x}}}\right)}+{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}$$
This will be solved by substitution method,
so assuming $$\displaystyle{t}={e}^{{{x}}}$$
differentiating,
$$\displaystyle{\frac{{{\left.{d}{t}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{d}{e}^{{{x}}}}}{{{\left.{d}{x}\right.}}}}$$
$$\displaystyle{\frac{{{\left.{d}{t}\right.}}}{{{\left.{d}{x}\right.}}}}={e}^{{{x}}}$$
$$\displaystyle{\left.{d}{t}\right.}={e}^{{{x}}}{\left.{d}{x}\right.}$$
Step 2
Substituting above values in the integral,
$$\displaystyle\int{\frac{{{e}^{{{x}}}}}{{{\left({e}^{{{x}}}\right)}^{{{2}}}+{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}=\int{\frac{{{\left.{d}{t}\right.}}}{{{t}^{{{2}}}+{2}{t}+{1}}}}$$
$$\displaystyle=\int{\frac{{{\left.{d}{t}\right.}}}{{{\left({t}\right)}^{{{2}}}+{2}{\left({t}\right)}{\left({1}\right)}+{\left({1}\right)}^{{{2}}}}}}$$
$$\displaystyle=\int{\frac{{{\left.{d}{t}\right.}}}{{{\left({t}+{1}\right)}^{{{2}}}}}}$$
$$\displaystyle=-{\frac{{{1}}}{{{t}+{1}}}}+{C}$$ (since $$\displaystyle\int{\frac{{{1}}}{{{\left({x}+{a}\right)}^{{{2}}}}}}=-{\frac{{{1}}}{{{x}+{a}}}}+{C}$$)
Where, C is an arbitrary constant.
Hence, value of integral is $$\displaystyle-{\frac{{{1}}}{{{e}^{{{x}}}+{1}}}}+{C}$$.
###### Not exactly what you’re looking for?
Barbara Meeker
It is required to calculate:
$$\displaystyle\int{\frac{{{e}^{{{x}}}}}{{{e}^{{{2}{x}}}+{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}$$
Substitution $$\displaystyle{u}={e}^{{{x}}}\Rightarrow{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}={e}^{{{x}}}\Rightarrow{\left.{d}{x}\right.}={e}^{{-{x}}}{d}{u}$$, we use:
$$\displaystyle{e}^{{{2}{x}}}={u}^{{{2}}}$$
$$\displaystyle=\int{\frac{{{1}}}{{{u}^{{{2}}}+{2}{u}+{1}}}}{d}{u}$$
Let us factorize:
$$\displaystyle=\int{\frac{{{1}}}{{{\left({u}+{1}\right)}^{{{2}}}}}}{d}{u}$$
Substitution $$\displaystyle{v}={u}+{1}\Rightarrow{\frac{{{d}{v}}}{{{d}{u}}}}={1}\Rightarrow{d}{u}={d}{v}:$$
$$\displaystyle=\int{\frac{{{1}}}{{{v}^{{{2}}}}}}{d}{v}$$
Integral of a power function:
$$\displaystyle\int{v}^{{{n}}}{d}{v}={\frac{{{v}^{{{n}+{1}}}}}{{{n}+{1}}}}$$ 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:
$$\displaystyle\int{\frac{{{e}^{{{x}}}}}{{{e}^{{{2}{x}}}+{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}$$
$$\displaystyle=-{\frac{{{1}}}{{{e}^{{{x}}}+{1}}}}+{C}$$
nick1337

$$\int \frac{e^{x}}{e^{2x}+2e^{x}+1}dx$$
Let us put the expression exp (x) under the sign of the differential, ie:
$$e^{x}*dx=d(e^{x}), t=e^{x}$$
Then the original integral can be written as follows:
$$\int \frac{1}{t^{2}+2*t+1}*dt$$
$$\int \frac{1}{x^{2}+2*x+1}*dx$$
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:
$$\int \frac{1}{(x+1)^{2}}*dx=-\frac{1}{x+1}$$
NSK
$$-\frac{1}{x+1}+C$$
To write down the final answer, it remains to substitute exp (x) instead of t.
$$-\frac{1}{e^{x}+1}+C$$