Chapter G, Problem 43E

(a)

To determine

To Calculate:

The partial fraction decomposition of the function.

Answer to Problem 43E

The partial fraction decomposition of the $f\left(x\right)=\frac{4x^3-27x^2+5x-32}{30x^5-13x^4+50x^3-286x^2-299x-70}$ is $f\left(x\right)=\frac{24110}{4879}\frac{1}{\left(5x+2\right)}-\frac{668}{3232}\frac{1}{\left(2x+1\right)}-\frac{9438}{80155}\frac{1}{\left(3x-7\right)}+\frac{1}{260015}\frac{22098x+48935}{x^2+x+5}$

Explanation of Solution

Given Information:

  $f\left(x\right)=\frac{4x^3-27x^2+5x-32}{30x^5-13x^4+50x^3-286x^2-299x-70}$

Calculation:

Consider the function,

  $f\left(x\right)=\frac{4x^3-27x^2+5x-32}{30x^5-13x^4+50x^3-286x^2-299x-70}$

Now, find the partial fraction decomposition using CAS.

So, maple input command:

  $f:=\left(4x^3-27x^2+5x-32\right)/\left(30x^5-13x^4+50x^3-286x^2-299x-70\right)$

Now, maple input and output as follows:

  $\begin{array}{l}>f:\frac{4x^3-27x^2+5x-32}{30x^5-13x^4+50x^3-286x^2-299x-70}\\ f:\frac{4x^3-27x^2+5x-32}{30x^5-13x^4+50x^3-286x^2-299x-70}\end{array}$

To display the partial fraction decomposition for the given function, put the maple command as follows,

  $\begin{array}{l}>convert\left(f,parfrac,x\right)\\ -\frac{9438}{80155\left(3x-7\right)}+\frac{1}{260015}\frac{22098x+48935}{x^2+x+5}-\frac{668}{323\left(2x+1\right)}+\frac{24110}{4879\left(5x+2\right)}\end{array}$

Therefore, partial fraction decomposition of the given function is,

  $f\left(x\right)=\frac{24110}{4879}\frac{1}{\left(5x+2\right)}-\frac{668}{3232}\frac{1}{\left(2x+1\right)}-\frac{9438}{80155}\frac{1}{\left(3x-7\right)}+\frac{1}{260015}\frac{22098x+48935}{x^2+x+5}$

(b)

To determine

To Calculate:

The ${\displaystyle \int f\left(x\right)dx}$ by using part (a) and compare with CAS to integrate $f$ directly.

Answer to Problem 43E

The ${\displaystyle \int f\left(x\right)dx}$ is

  $\begin{array}{l}\frac{4822}{4879}\ln \left(5x+2\right)-\frac{334}{323}\ln \left(2x+1\right)+\frac{11049}{260015}\ln \left(x^2+x+5\right)+\frac{3988}{260015}\sqrt{19}\arctan \left(\frac{1}{19}\left(2x+1\right)\sqrt{19}\right)\\ -\frac{3146}{80155}\ln \left(3x-7\right)\end{array}$

Explanation of Solution

Given Information:

  $f\left(x\right)=\frac{4x^3-27x^2+5x-32}{30x^5-13x^4+50x^3-286x^2-299x-70}$

Calculation:

  ${\displaystyle \int f\left(x\right)dx}$

  $={\displaystyle \int \frac{4x^3-27x^2+5x-32}{30x^5-13x^4+50x^3-286x^2-299x-70}dx}$

  $={\displaystyle \int \left[\frac{24110}{4879}\frac{1}{\left(5x+2\right)}-\frac{668}{3232}\frac{1}{\left(2x+1\right)}-\frac{9438}{80155}\frac{1}{\left(3x-7\right)}+\frac{1}{260015}\frac{22098x+48935}{x^2+x+5}\right]dx}$

  $=\frac{24110}{4879}{\displaystyle \int \frac{1}{\left(5x+2\right)}}dx-\frac{668}{323}{\displaystyle \int \frac{1}{\left(2x+1\right)}}dx-\frac{9438}{80155}{\displaystyle \int \frac{1}{\left(3x-7\right)}dx+\frac{1}{260015}}{\displaystyle \int \frac{22098x+48935}{x^2+x+5}}dx$

  $=\frac{24110}{4879}\cdot \frac{1}{5}{\displaystyle \int \frac{5}{\left(5x+2\right)}}dx-\frac{668}{323}\cdot \frac{1}{2}{\displaystyle \int \frac{2}{\left(2x+1\right)}}dx-\frac{9438}{80155}\cdot \frac{1}{3}{\displaystyle \int \frac{3}{\left(3x-7\right)}}dx$

  $+\frac{1}{260015}{\displaystyle \int \left(\frac{22098x}{x^2+x+5}+\frac{11049}{x^2+x+5}-\frac{11049}{x^2+x+5}+\frac{48935}{x^2+x+5}\right)}dx$

  ${\displaystyle \int f\left(x\right)}dx=\frac{24110}{4879}\cdot \frac{1}{5}{\displaystyle \int \frac{5}{\left(5x+2\right)}}dx-\frac{668}{323}\cdot \frac{1}{2}{\displaystyle \int \frac{2}{\left(2x+1\right)}}dx-\frac{9438}{80155}\cdot \frac{1}{3}{\displaystyle \int \frac{3}{\left(3x-7\right)}}dx$

  $+\frac{1}{260015}{\displaystyle \int \left(\frac{22098x+11049}{x^2+x+5}+\frac{37886}{x^2+x+5}\right)}dx$

  $=\frac{24110}{4879}\cdot \frac{1}{5}{\displaystyle \int \frac{5}{\left(5x+2\right)}}dx-\frac{668}{323}\cdot \frac{1}{2}{\displaystyle \int \frac{2}{\left(2x+1\right)}}dx-\frac{9438}{80155}\cdot \frac{1}{3}{\displaystyle \int \frac{3}{\left(3x-7\right)}}dx+$

  $\frac{1}{260015}{\displaystyle \int \left(\frac{22098x+11049}{x^2+x+5}\right)dx}+\frac{1}{260015}{\displaystyle \int \frac{37886}{x^2+x+5}}dx$

  $=\frac{24110}{4879}\cdot \frac{1}{5}{\displaystyle \int \frac{5}{\left(5x+2\right)}}dx-\frac{668}{323}\cdot \frac{1}{2}{\displaystyle \int \frac{2}{\left(2x+1\right)}}dx-\frac{9438}{80155}\cdot \frac{1}{3}{\displaystyle \int \frac{3}{\left(3x-7\right)}}dx+$

  $\frac{11049}{260015}{\displaystyle \int \left(\frac{2x+1}{x^2+x+5}\right)}dx+\frac{37886}{260015}{\displaystyle \int \frac{1}{x^2+x+5}}dx$

  $=\frac{24110}{4879}\cdot \frac{1}{5}{\displaystyle \int \frac{5}{\left(5x+2\right)}}dx-\frac{668}{323}\cdot \frac{1}{2}{\displaystyle \int \frac{2}{\left(2x+1\right)}}dx-\frac{9438}{80155}\cdot \frac{1}{3}{\displaystyle \int \frac{3}{\left(3x-7\right)}}dx+$

  $\frac{11049}{260015}{\displaystyle \int \left(\frac{2x+1}{x^2+x+5}\right)}dx+\frac{37886}{260015}{\displaystyle \int \frac{1}{{\left(x+\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{19}}{2}\right)}^2}}dx$

Recalling that,

  ${\displaystyle \int \frac{f\text{'}\left(x\right)}{f\left(x\right)}}dx=\ln \left|f\left(x\right)\right|+C$ and ${\displaystyle \int \frac{1}{x^2+a^2}}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)$

  $\begin{array}{l}{\displaystyle \int f\left(x\right)}dx=\frac{4822}{4879}\ln \left|5x+2\right|-\frac{334}{323}\ln \left|2x+1\right|-\frac{3146}{80155}\ln \left|3x-7\right|\\ +\frac{11049}{260015}\ln \left|x^2+x+5\right|+\frac{37886}{260015}\cdot \frac{1}{\sqrt{\frac{19}{2}}}{\tan }^{-1}\left(\frac{\frac{2x+1}{2}}{\frac{\sqrt{19}}{2}}\right)\\ =\frac{4822}{4879}\ln \left|5x+2\right|-\frac{334}{323}\ln \left|2x+1\right|-\frac{3146}{80155}\ln \left|3x-7\right|\\ +\frac{11049}{260015}\ln \left|x^2+x+5\right|+\frac{75772}{260015\sqrt{9}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{19}}\right)+C\end{array}$

Therefore,

  $\begin{array}{l}{\displaystyle \int f\left(x\right)dx}=\frac{4822}{4879}\ln \left|5x+2\right|-\frac{334}{323}\ln \left|2x+1\right|-\frac{3146}{80,155}\ln \left|3x-7\right|+\\ \frac{11049}{260,015}\ln \left|x^2+x+5\right|+\frac{75,772}{260,015\sqrt{19}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{19}}\right)+C\end{array}$

By using CAS,

  $\begin{array}{l}>{\displaystyle \int \left(\frac{4x^3-27x^2+5x-32}{30x^5-13x^4+50x^3-286x^2-299x-70}\right)}dx\\ \frac{4822}{4879}\ln \left(5x+2\right)-\frac{334}{323}\ln \left(2x+1\right)+\frac{11049}{260015}\ln \left(x^2+x+5\right)+\\ \frac{3988}{260015}\sqrt{19}\arctan \left(\frac{1}{19}\left(2x+1\right)\sqrt{19}\right)-\frac{3146}{80155}\ln \left(3x-7\right)\end{array}$

As from the above results, by hand and CAS, observe that the CAS omits the absolute sign and the constant of integration.