Question

I need help in partial fraction!! With simple explaination would be nice!

Answer 1

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}$

The degree of the numerator (4) is larger than the degree of the denominator (3), so first you need to divide. (Added screenshot of long division procedure.)

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x-\dfrac{4x^2-17x+10}{x(x^2-3)}$

Now the second term can be decomposed into partial fractions.

$\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1}x+\dfrac{r_2x+r_3}{x^2-3}$
$\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1(x^2-3)+x(r_2x+r_3)}{x(x^2-3)}$
$4x^2-17x+10=r_1(x^2-3)+x(r_2x+r_3)$
$4x^2-17x+10=(r_1+r_2)x^2+r_3x-3r_1$
$\implies\begin{cases}r_1+r_2=4\\r_3=-17\\-3r_1=10\end{cases}\implies r_1=-\dfrac{10}3,r_2=\dfrac{22}3,r_3=-17=-\dfrac{51}3$
$\implies\dfrac{4x^2-17x+10}{x(x^2-3)}=-\dfrac{10}{3x}+\dfrac{22x-51}{x^2-3}$

So

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x+\dfrac{10}{3x}-\dfrac{22x-51}{x^2-3}$