Question

Resolve into partial fractions:
$\frac{x^4+3x^3-3x^2+10}{{(x+1)}^2(x-3)}$

Answer 1

Resolve into partial fractions

$\frac{x^4+3x^3-3x^2+10}{(x+1{)}^2(x-3)}$

$=x-2-\frac{7x-4}{(x+1{)}^2(x-3)}$

Now assume,

$\frac{7x-4}{(x+1{)}^2(x-3)}=\frac{A}{(x+1)}+\frac{B}{(x+1{)}^2}+\frac{C}{(x-3)}$

After Simplifying this, we get

$A(x+1)(x-3)+B(x-3)+C(x+1{)}^2=7x-4$

On solving for $A,B$ and $C$, we get

$A=\frac{-17}{16},B=\frac{11}{4},C=\frac{17}{16}$

$\therefore$ $\frac{x^4+3x^3-3x^2+10}{(x+1{)}^2(x-3)}$

$=x-2-\frac{-17}{16(x+1)}+\frac{11}{4(x+1{)}^2}+\frac{17}{16(x-3)}$