Question

Find the general term of the following expressions when expanded in ascending powers of $x$.
$\frac{2x+1}{(x-1)(x^2+1)}$

Answer 1

To find the general term of the following expressions when expanded in ascending power of x .

$\frac{2x+1}{(x-1)(x^2+1)}$

Here , we have

$\frac{2x+1}{(x-1)(x^2+1)}$

The Expression can be written as

$\frac{2x+1}{(x-1)(x^2+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{x^2+1}$

On Solving for $A,B$ we have

$A=\frac{3}{2},B=\frac{-3}{2},C=\frac{1}{2}$

Therefore , The Above Expression as

$\frac{3}{2(x-1)}+\frac{1-3x}{2(x^2+1)}$

Now the General Term can be written as

$\frac{1}{2}\left\{\right(-1{)}^{r/2}-3\}{x}^r$ , when r is even

$-\frac{3}{2}\{1+(-1{)}^{\frac{r-1}{r}}\}{x}^r$ , when r is odd