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

Irvin Dukes 2021-12-20 Answered
Evaluate the following integrals.
\(\displaystyle\int{\frac{{{e}^{{{x}}}}}{{{e}^{{{2}{x}}}+{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

kalupunangh
Answered 2021-12-21 Author has 457 answers
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?
Ask My Question
0
 
Barbara Meeker
Answered 2021-12-22 Author has 6441 answers
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:
\(\displaystyle\int{\frac{{{e}^{{{x}}}}}{{{e}^{{{2}{x}}}+{2}{e}^{{{x}}}+{1}}}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{\frac{{{1}}}{{{e}^{{{x}}}+{1}}}}+{C}\)
0
nick1337
Answered 2021-12-28 Author has 10701 answers

\(\int \frac{e^{x}}{e^{2x}+2e^{x}+1}dx\)
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
Answer:
\(-\frac{1}{x+1}+C\)
To write down the final answer, it remains to substitute exp (x) instead of t.
\(-\frac{1}{e^{x}+1}+C\)

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more
...