Chapter 5.7, Problem 25E

To determine

To calculate: The value of given integral

Answer to Problem 25E

The value of the given integral is $\ln \left|x-1\right|-\frac{1}{2}\ln \left|x^2+9\right|-\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \frac{10}{\left(x-1\right)\left(x^2+9\right)}}dx$

Formula used:

  ${\displaystyle \int \frac{1}{x}dx=\ln \left|x\right|+C}$

  ${\displaystyle \int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)}$

Calculation:

The partial function expansion of the integrand is,

  $\frac{10}{\left(x-1\right)\left(x^2+9\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}$

Here, $A,B\text{ and C}$ are constants. To find the value of $A,B\text{ and C}$ multiply both sides of the equation by $\left(x-1\right)\left(x^2+9\right)$

  $\begin{array}{l}10=A\left(x^2+9\right)+\left(Bx+C\right)\left(x-1\right)\\ 10=A{x}^2+9A+B{x}^2-Bx+Cx-C\\ 10=x^2\left(A+B\right)+x\left(C-B\right)+9A-C\end{array}$

The coefficients of $x$ must be equal and the constant terms are also equal. So

  $A+B=0,C-B=0\text{ and }9A-C=10$

Solve this system of linear equation for $A,B\text{ and C}$ , by use of simple substitution

  $\begin{array}{l}0=A+B\Rightarrow A=-B\\ 0=C-B\Rightarrow C=B\Rightarrow C=-A\\ 10=9A-C\Rightarrow 10=-A+A\end{array}$

So, $A=1,B=-1\text{ and }C=-1$

Each of the resultants partial function is easy to integrate. Evaluate integral on the right side of the equality. Use the substitution $u=x-1\text{ and }v=x^2+9$ respectively. So,

  ${\displaystyle \int \frac{10}{\left(x-1\right)\left(x^2+9\right)}}dx={\displaystyle \int \frac{1}{x-1}dx+{\displaystyle \int \frac{-x-1}{x^2+9}}}dx$

Integrate the function to get the final result.

  $\begin{array}{l}{\displaystyle \int \frac{10}{\left(x-1\right)\left(x^2+9\right)}}dx\\ ={\displaystyle \int \frac{1}{x-1}dx+{\displaystyle \int \frac{-x-1}{x^2+9}}}dx\\ ={\displaystyle \int \frac{1}{x-1}dx-{\displaystyle \int \frac{x+1}{x^2+9}dx}}\\ ={\displaystyle \int \frac{1}{x-1}dx}-{\displaystyle \int \frac{x}{x^2+9}dx-}{\displaystyle \int \frac{1}{x^2+9}dx}\\ ={\displaystyle \int \frac{1}{u}du}-\frac{1}{2}{\displaystyle \int \frac{1}{v}dv-{\displaystyle \int \frac{1}{x^2+9}dx}}\\ =\ln \left|u\right|-\frac{1}{2}\ln \left|v\right|-\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)\left[\text{formula:}{\displaystyle \int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)}\right]\\ =\ln \left|x-1\right|-\frac{1}{2}\ln \left|x^2+9\right|-\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+C\end{array}$

Therefore, the value of the given integral is $\ln \left|x-1\right|-\frac{1}{2}\ln \left|x^2+9\right|-\frac{1}{3}{\tan }^{-1}\left(\frac{x}{3}\right)+C$