Question

Evaluate the integral
∫(ax)/(x^2−bx)dx,
Hint: The a and b are constants. Treat them as you would π or e.

Answer 1

\[\begin{align*}\int\frac{ax}{x^2-bx}~dx&=\frac{a}{2}\left(\int\frac{2x}{x^2-bx}~dx\right)\\
&=\frac{a}{2}\left(\int\frac{2x-b}{x^2-bx}~dx+\int\frac{b}{x^2-bx}~dx\right)
\end{align*}\]
The first integral uses a substitution: \(u=x^2-bx~~\Rightarrow~~du=(2x-b)~dx\).

For the second integral, I would complete the square in the denominator, so that you have
\[\frac{b}{x^2-bx}=\frac{b}{x^2-bx+\frac{b^2}{4}-\frac{b^2}{4}}=\frac{b}{\left(x-\frac{b}{2}\right)^2-\frac{b^2}{4}}\]
So you have
\[\frac{a}{2}\left(\int\frac{du}{u}+\int\frac{b}{\left(x-\frac{b}{2}\right)^2-\frac{b^2}{4}}~dx\right)\]
Finally, a trig sub should do the trick: \(x-\dfrac{b}{2}=\dfrac{b}{2}\sec t~~\Rightarrow~~dx=\dfrac{b}{2}\sec t\tan t~dt\):
\[\frac{a}{2}\left(\int\frac{du}{u}+\int\frac{b}{\left(\frac{b}{2}\sec t\right)^2-\frac{b^2}{4}}~\left(\frac{b}{2}\sec t\tan t~dt\right)\right)\]
\[\frac{a}{2}\left(\int\frac{du}{u}+2\int\frac{\sec t\tan t}{\sec^2 t-1}~dt\right)\]
\[\frac{a}{2}\left(\int\frac{du}{u}+2\int\csc t~dt\right)\]