Chapter G, Problem 37E

To determine

To calculate:

The integral value of the following functions by using substitution method:

  ${\displaystyle \int \frac{e^{2x}}{e^{2x}+3e^x+2}dx}$

Answer to Problem 37E

The integral value of ${\displaystyle \int \frac{e^{2x}}{e^{2x}+3e^x+2}dx}$ is $\begin{array}{l}\frac{\ln \left(u^2+3u+2\right)}{2}+\frac{3\ln \left(u+2\right)}{2}-\frac{3\ln \left(e^x+1\right)}{2}+C\end{array}$ .

Explanation of Solution

Given information:

The integral function:

  ${\displaystyle \int \frac{e^{2x}}{e^{2x}+3e^x+2}dx}$

Calculation:

The integral function is given that:

  ${\displaystyle \int \frac{e^{2x}}{e^{2x}+3e^x+2}dx}$

Substitute the value:

  $\begin{array}{l}u=e^x\\ \frac{d}{dx}\left(u\right)=\frac{d}{dx}\left(e^x\right)\\ dx=e^{-x}du\\ e^{2x}=u^2\end{array}$

  $={\displaystyle \int \frac{u}{u^2+3u+2}du}$

Write $u$ as $\frac{1}{2}\left(2u+3\right)-\frac{3}{2}$ and then split:

  $={\displaystyle \int \left(\frac{2u+3}{2\left(u^2+3u+2\right)}-\frac{3}{2\left(u^2+3u+2\right)}\right)}du$

Apply linearity form:

  $=\frac{1}{2}{\displaystyle \int \frac{2u+3}{u^2+3u+2}du-\frac{3}{2}{\displaystyle \int \frac{1}{u^2+3u+2}}}du$   ...(1)

Simplify the first term of above right hand side of equation (1)

  $\frac{2u+3}{u^2+3u+2}du$

Substitute the value:

  $\begin{array}{l}v=u^2+3u+2\\ \frac{d}{dx}\left(v\right)=\frac{d}{dx}\left(u^2+3u+2\right)\\ \frac{dv}{du}=2u+3\\ du=\frac{1}{2u+3}dv\\ ={\displaystyle \int \frac{1}{v}dv}\end{array}$

Know that, this is the standard integral.

  ${\displaystyle \int \frac{1}{v}dv=\ln \left(v\right)}$

Substitute the value of $v$ ,

  ${\displaystyle \int \frac{1}{v}dv=\ln \left(u^2+3u+2\right)}$

Simplify the second term of above right hand side of equation (1)

  ${\displaystyle \int \frac{1}{u^2+3u+2}du}$

Factor the equation denominator:

  ${\displaystyle \int \frac{1}{u^2+3u+2}du}={\displaystyle \int \frac{1}{\left(u+1\right)\left(u+2\right)}du}$

By this function, need to perform partial fraction decomposition:

  ${\displaystyle \int \frac{1}{u^2+3u+2}du}={\displaystyle \int \left(\frac{1}{u+1}-\frac{1}{u+2}\right)du}$

Apply the linearity:

  ${\displaystyle \int \frac{1}{u^2+3u+2}du}={\displaystyle \int \frac{1}{u+1}du-}{\displaystyle \int \frac{1}{u+2}du}$  ...(2)

Simplify the first term of above right hand side of equation (2)

  ${\displaystyle \int \frac{1}{u+1}du}$

Substitute the value:

  $\begin{array}{l}v=u+1\\ \frac{dv}{du}=1\\ du=dv\\ ={\displaystyle \int \frac{1}{v}dv}\end{array}$

Know that,

  ${\displaystyle \int \frac{1}{v}dv=\ln \left(v\right)}$

Put the value of $v$ ,

  ${\displaystyle \int \frac{1}{v}dv=\ln \left(u+1\right)}$

Simplify the second term of above right hand side of equation (2)

  ${\displaystyle \int \frac{1}{u+2}du}$

Substitute the value:

  $\begin{array}{l}v=u+2\\ \frac{dv}{du}=1\\ du=dv\\ ={\displaystyle \int \frac{1}{v}dv}\end{array}$

Know that,

  ${\displaystyle \int \frac{1}{v}dv=\ln \left(v\right)}$

Put the value of $v$ ,

  ${\displaystyle \int \frac{1}{v}dv=\ln \left(u+2\right)}$

Put all the values in equation (2)

  ${\displaystyle \int \frac{1}{u^2+3u+2}du}={\displaystyle \int \frac{1}{u+1}du-}{\displaystyle \int \frac{1}{u+2}du}$

  ${\displaystyle \int \frac{1}{u^2+3u+2}du}=\ln \left(u+1\right)-\ln \left(u+2\right)$

Then, plug all the values in equation (1),

  ${\displaystyle \int \frac{e^{2x}}{e^{2x}+3e^x+2}dx}=\frac{1}{2}{\displaystyle \int \frac{2u+3}{u^2+3u+2}du-\frac{3}{2}{\displaystyle \int \frac{1}{u^2+3u+2}}}du$

  $\begin{array}{l}{\displaystyle \int \frac{e^{2x}}{e^{2x}+3e^x+2}dx}=\frac{\ln \left(u^2+3u+2\right)}{2}+\frac{3\ln \left(u+2\right)}{2}-\frac{3\ln \left(e^x+1\right)}{2}+C\end{array}$