Chapter G, Problem 18E

To determine

To calculate:

The integral value of the following function:

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

Answer to Problem 18E

The integral value of ${\displaystyle \int \frac{x^2+2x-1}{x^3-x}}dx$ is $\ln \left|\frac{x\left(x-1\right)}{x+1}\right|+C$ .

Explanation of Solution

Given information:

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

Calculation:

The given equation is:

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

Use the partial factor decomposition here:

  $\begin{array}{l}\frac{x^2+2x-1}{x^3-x}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\\ x^2+2x-1=A\left(x-1\right)\left(x+1\right)+Bx\left(x+1\right)+Cx\left(x-1\right)\\ x^2+2x-1=A{x}^2-A+B{x}^2+Bx+C{x}^2-Cx\end{array}$

On comparing, get the equations as:

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

Put $A=1$ in

  $\begin{array}{l}A+B+C=1\\ 1+B+C=1\\ B+C=0\end{array}$

Solve two eqns:

  $\begin{array}{l}B+C=0\\ B-C=2\\ -------\\ 2B=2\\ B=1\end{array}$

So,

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

Hence

  $\begin{array}{l}{\displaystyle \int \frac{x^2+2x-1}{x^3-x}}dx={\displaystyle \int \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}}dx\\ {\displaystyle \int \frac{x^2+2x-1}{x^3-x}}dx={\displaystyle \int \left(\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x+1}\right)}dx\\ \left[\because {\displaystyle \int \left(\frac{1}{x}\right)}dx=\ln x\right]\\ {\displaystyle \int \frac{x^2+2x-1}{x^3-x}}dx=\ln x+\ln \left(x-1\right)-\ln \left(x+1\right)+C\\ {\displaystyle \int \frac{x^2+2x-1}{x^3-x}}dx=\ln \left|\frac{x\left(x-1\right)}{x+1}\right|+C\\ \end{array}$