Chapter G, Problem 19E

To determine

To calculate: ${\displaystyle {\int }_{}^{}\frac{1}{{\left(x+5\right)}^2\left(x-1\right)}dx}$

Answer to Problem 19E

  $\frac{\ln \left(\left|x-1\right|\right)}{36}-\frac{\ln \left(\left|x+5\right|\right)}{36}+\frac{1}{6\left(x+5\right)}+C$

Explanation of Solution

Given information :

  ${\displaystyle {\int }_{}^{}\frac{1}{{\left(x+5\right)}^2\left(x-1\right)}dx}$

Formula used :

Partial fraction decomposition:

  $\frac{p{x}^2+qx+r}{{\left(x-a\right)}^2\left(x-b\right)}=\frac{A}{x-a}+\frac{B}{{\left(x-a\right)}^2}+\frac{C}{x-b}$

  ${\displaystyle {\int }_{}^{}\frac{1}{x}}dx=\ln \left(x\right)$

Power rule: ${\displaystyle {\int }_{}^{}{a}^{n}dx}=\frac{a^{n+1}}{n+1},n\ne -1$

Calculation:

  ${\displaystyle {\int }_{}^{}\frac{1}{{\left(x+5\right)}^2\left(x-1\right)}dx}$

Apply partial fraction decomposition:

  $\frac{1}{{\left(x+5\right)}^2\left(x-1\right)}=\frac{A}{x+5}+\frac{B}{{\left(x+5\right)}^2}+\frac{C}{x-1}$

Write the right-hand side as a single fraction:

  $\frac{1}{{\left(x+5\right)}^2\left(x-1\right)}=\frac{A\left(x+5\right)\left(x-1\right)+\left(x-1\right)B+{\left(x+5\right)}^{2}C}{{\left(x+5\right)}^2\left(x-1\right)}$

The denominators are equal, so we require the equality of the numerators:

  $1=A\left(x+5\right)\left(x-1\right)+\left(x-1\right)B+{\left(x+5\right)}^{2}C$

Expand the right-hand side:

  $1=x^{2}A+x^{2}C+4xA+xB+10xC-5A-B+25C$

Collect up the like terms:

  $1=x^2\left(A+C\right)+x\left(4A+B+10C\right)-5A-B+25C$

The coefficients near the like terms should be equal, so the following system is obtained:

  $\{\begin{array}{l}A+C=0\Rightarrow A=-C\\ 4A+B+10C=0\Rightarrow 4\left(-C\right)+B+10C=0\Rightarrow B+6C=0\Rightarrow B=-6C\mathrm{.....}\left(1\right)\\ -5A-B+25C=1\Rightarrow -5\left(-C\right)-B+25C=1\Rightarrow -B+30C=1\mathrm{.....}\left(2\right)\end{array}$

By (1) and (2), we get:

  $\begin{array}{l}-\left(-6C\right)+30C=1\\ \Rightarrow 6C+60C=1\\ \Rightarrow \boxed{C=\frac{1}{36}}\end{array}$

Plug the value of ‘C’ in equation (1), we get:

  $\begin{array}{l}B=-6\times \frac{1}{36}\\ \Rightarrow \boxed{B=-\frac{1}{6}}\end{array}$

  $A=-C\Rightarrow \boxed{A=-\frac{1}{36}}$

$\frac{1}{{\left(x+5\right)}^2\left(x-1\right)}=\frac{-\frac{1}{36}}{x+5}+\frac{-\frac{1}{6}}{{\left(x+5\right)}^2}+\frac{\frac{1}{36}}{x-1}$

$\Rightarrow {\displaystyle {\int }_{}^{}\frac{1}{{\left(x+5\right)}^2\left(x-1\right)}}dx={\displaystyle {\int }_{}^{}\left(\frac{1}{36\left(x-1\right)}-\frac{1}{36\left(x+5\right)}-\frac{1}{6{\left(x+5\right)}^2}\right)dx}$

$=\frac{1}{36}{\displaystyle {\int }_{}^{}\frac{1}{\left(x-1\right)}dx}-\frac{1}{36}{\displaystyle {\int }_{}^{}\frac{1}{\left(x+5\right)}}dx-\frac{1}{6}{\displaystyle {\int }_{}^{}\frac{1}{{\left(x+5\right)}^2}dx}$

$\begin{array}{l}\text{Apply substitution: }u=x-1\Rightarrow \frac{du}{dx}=1\Rightarrow dx=du\\ \text{Apply substitution: }v=x+5\Rightarrow \frac{dv}{dx}=1\Rightarrow dx=dv\\ \text{Apply substitution: w}=x+5\Rightarrow \frac{dw}{dx}=1\Rightarrow dx=dw\end{array}$

$=\frac{1}{36}{\displaystyle {\int }_{}^{}\frac{1}{u}du}-\frac{1}{36}{\displaystyle {\int }_{}^{}\frac{1}{v}}dv-\frac{1}{6}{\displaystyle {\int }_{}^{}\frac{1}{{\left(w\right)}^2}dw}$

$=\frac{1}{36}{\displaystyle {\int }_{}^{}\frac{1}{u}du}-\frac{1}{36}{\displaystyle {\int }_{}^{}\frac{1}{v}}dv-\frac{1}{6}{\displaystyle {\int }_{}^{}{w}^{-2}dw}\left(\because a^{-n}=\frac{1}{a^n}\right)$

$=\frac{\ln \left(u\right)}{36}-\frac{\ln \left(v\right)}{36}-\frac{1}{6}\left[\frac{w^{-2+1}}{-2+1}\right]\left(\because \text{Power rule}\right)$

$=\frac{\ln \left(u\right)}{36}-\frac{\ln \left(v\right)}{36}+\frac{1}{6w}\left(\because a^{-n}=\frac{1}{a^n}\right)$

$\begin{array}{l}\text{Substitute back: }u=x-1,v=x+5,w=x+5\\ \Rightarrow {\displaystyle {\int }_{}^{}\frac{1}{{\left(x+5\right)}^2\left(x-1\right)}}dx=\frac{\ln \left(\left|x-1\right|\right)}{36}-\frac{\ln \left(\left|x+5\right|\right)}{36}+\frac{1}{6\left(x+5\right)}+C\end{array}$