Question

Resolve into partial fractions:
$\frac{7x-1}{1-5x+6x^2}$

Answer 1

The table give general form of the partial fractions :

$\frac{7x-1}{1-5x+6x^2}=\frac{1}{6}\left(\frac{7x-1}{\frac{1}{6}-\frac{5x}{6}+x^2}\right)$

$\Rightarrow \frac{1}{6}\left(\frac{7x-1}{(x-\frac{1}{2})(x-\frac{1}{3})}\right)$ $=\frac{A}{x-\frac{1}{2}}+\frac{B}{x-\frac{1}{3}}$

So, $\frac{1}{6}(7x-1)=A(x-\frac{1}{3})+B(x-\frac{1}{2})$

For $x=\frac{1}{3}\Rightarrow \frac{1}{6}(\frac{7}{3}-1)=B(\frac{1}{3}-\frac{1}{2})$

$\Rightarrow \frac{4}{18}=-\frac{B}{6}$

$\Rightarrow B=-\frac{4}{3}$

For $x=\frac{1}{2}\Rightarrow \frac{1}{6}(\frac{7}{2}-1)=A(\frac{1}{2}-\frac{1}{3})$

$\Rightarrow \frac{5}{12}=\frac{A}{6}$

$\Rightarrow A=\frac{5}{2}$

Therefore, $\frac{7x-1}{1-5x+6x^2}=\frac{5}{2}\left(\frac{1}{x-\frac{1}{2}}\right)-\frac{4}{3}\left(\frac{1}{x-\frac{1}{3}}\right)$