Chapter 7.5, Problem 21E

To determine

To find the integral without using partial fractions technique.

Answer to Problem 21E

The value of ${\displaystyle \int \frac{x^2+x-1}{x^2-x}}dx=x-\log \left|\frac{x-1}{x}\right|+C$

Explanation of Solution

Given information:

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

Calculation:

  $\begin{array}{l}{\displaystyle \int \frac{x^2+x-1}{x^2-x}}dx\\ \text{given expression }\frac{x^2+x-1}{x^2-x}\text{can be written as}\\ \frac{x^2+x-1}{x^2-x}=1-\frac{1}{x^2-x}=1-\frac{1}{x^2-x+{\left(1/2\right)}^2-{\left(1/2\right)}^2}=1-\frac{1}{{\left(x-\left(1/2\right)\right)}^2-{\left(1/2\right)}^2}\\ {\displaystyle \int \frac{x^2+x-1}{x^2-x}}dx={\displaystyle \int \left(1-\frac{1}{{\left(x-\left(1/2\right)\right)}^2-{\left(1/2\right)}^2}\right)}dx\\ =x-\frac{1}{2\times 1/2}\log \left|\frac{x-1/2-1/2}{x-1/2+1/2}\right|+C\\ =x-\log \left|\frac{x-1}{x}\right|+C,\text{ where }C\text{ is constant}\\ \therefore {\displaystyle \int \frac{x^2+x-1}{x^2-x}}dx=x-\log \left|\frac{x-1}{x}\right|+C\end{array}$

Therefore, the value of ${\displaystyle \int \frac{x^2+x-1}{x^2-x}}dx=x-\log \left|\frac{x-1}{x}\right|+C$