Chapter 5.7, Problem 24E

To determine

To calculate: The value of given integral

Answer to Problem 24E

The value of the given integral is $\ln \left(x-1\right)+\ln \left(x\right)-\ln \left(x-1\right)+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \frac{x^2+2x-1}{x^3-x}}dx$

Formula used:

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

Calculation:

The denominator $x^3-x$ is factored to $x\left(x+1\right)\left(x-1\right)$

Now, do the partial function expansion of the integrand

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

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 $x\left(x+1\right)\left(x-1\right)$

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

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

  $A+B+C=1,A-C=2\text{ and }-1=-B$

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

  $\begin{array}{l}B=1\\ A-C=2\Rightarrow A=2+C\Rightarrow A=2-1=1\\ 1=2+C+1+C\Rightarrow 1=3+2C\Rightarrow C=-1\end{array}$

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

Take the values found. Plot them back into the function that is found initially for the constants $A,B\text{ and C}$

  ${\displaystyle \int \frac{1}{x-1}+\frac{1}{x}-\frac{1}{x+1}}$

Integrate the function to get the final result

  $\ln \left(x-1\right)+\ln \left(x\right)-\ln \left(x-1\right)+C$

Therefore, the value of the given integral is $\ln \left(x-1\right)+\ln \left(x\right)-\ln \left(x-1\right)+C$