Question

Break $\frac{2x+1}{(x-1)(x^2+1)}$ into partial fractions.

Answer 1

Let, $\frac{2x+1}{(x-1)(x^2+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{(x^2+1)}$
$\frac{2x+1}{(x-1)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x-1)}{(x-1)(x^2-1)}$
$2x+1=A(x^2+1)+B(x^2-x)+C(x-1)$
Putting $x-1=0$ is $x=1$ in equation (2)
$2\left(1\right)+1=A{\left(1\right)}^2+1)+B({\left(1\right)}^2-1)+C(1-1)$
$3=2A+B\left(0\right)+C\left(0\right)$
$A=\frac{3}{2}$
Now, on comparing with coefficients of $x^2$ in equation(2)
$0=A+B$
On putting value of $A$
$0=\frac{3}{2}+B$
$B=\frac{-3}{2}$
Comparing with coeff. of $x$ in equation (2)
$2=-B+C$
putting value of $B$
$2=\frac{3}{2}+C$
$C=\frac{4-3}{2}\Rightarrow C=\frac{1}{2}$
putting values of $A,B,C$ in equation (1)
$\frac{2x+1}{(x-1)(x^2+1)}=\frac{3}{2}\frac{1}{(x-1)}+\frac{\left(\frac{-3}{2}\right)x+\left(\frac{1}{2}\right)}{x^2+1}$
$=\frac{\frac{(-3x+1)}{2}}{x^2+1}=\frac{1}{2}[\frac{3}{x-1}-\frac{3x-1}{x^2+1}]$