Chapter G, Problem 13E

To determine

To calculate:

The integral value of the following function:

  ${\displaystyle \int \frac{ax}{x^2-bx}}dx$

Answer to Problem 13E

The integral value of ${\displaystyle \int \frac{ax}{x^2-bx}}dx$ is $a\ln \left(x-b\right)+C$ .

Explanation of Solution

Given information:

  ${\displaystyle \int \frac{ax}{x^2-bx}}dx$

Calculation:

The given equation is:

  ${\displaystyle \int \frac{ax}{x^2-bx}}dx$

Simply it:

  $\begin{array}{l}{\displaystyle \int \frac{ax}{x^2-bx}}dx={\displaystyle \int \frac{ax}{x\left(x-b\right)}}dx\\ {\displaystyle \int \frac{ax}{x^2-bx}}dx={\displaystyle \int \frac{a}{\left(x-b\right)}}dx\end{array}$

Put: $x-b=u$

  $\begin{array}{l}{\displaystyle \int \frac{ax}{x^2-bx}}dx={\displaystyle \int \frac{a}{\left(x-b\right)}}dx\\ {\displaystyle \int \frac{ax}{x^2-bx}}dx={\displaystyle \int \frac{a}{u}}du\\ {\displaystyle \int \frac{ax}{x^2-bx}}dx=a\ln \left(u\right)\left[{\displaystyle \int \frac{1}{x}}dx=\ln x\right]\end{array}$

From eq. (1), undo the substitution as:

  $\begin{array}{l}{\displaystyle \int \frac{ax}{x^2-bx}}dx=a\ln \left(u\right)\\ {\displaystyle \int \frac{ax}{x^2-bx}}dx=a\ln \left(x-b\right)\end{array}$

Need to use the absolute value function to arguments of logarithm functions to extend the anti-derivative’s domain:

  ${\displaystyle \int \frac{ax}{x^2-bx}}dx=a\ln \left(x-b\right)+C$