Question

Resolve into partial fraction:
x+4/x(x+1)(x+2)​

Answer 1

Let,

$\displaystyle\longrightarrow\sf{\dfrac{x+4}{x(x+1)(x+2)}=\dfrac{A}{x}+\dfrac{B}{x+1}+\dfrac{C}{x+2}}$

$\displaystyle\longrightarrow\sf{\dfrac{x+4}{x(x+1)(x+2)}=\dfrac{A(x+1)(x+2)+Bx(x+2)+Cx(x+1)}{x(x+1)(x+2)}}$

Equating the numerators,

$\displaystyle\longrightarrow\sf{x+4=A(x^2+3x+2)+B(x^2+2x)+C(x^2+x)}$

$\displaystyle\longrightarrow\sf{x+4=Ax^2+3Ax+2A+Bx^2+2Bx+Cx^2+Cx}$

$\displaystyle\longrightarrow\sf{x+4=(A+B+C)x^2+(3A+2B+C)x+2A}$

From this,

$\displaystyle\longrightarrow\sf{2A=4\quad\implies\quad A=2}$

$\displaystyle\longrightarrow\sf{A+B+C=2+B+C=0\quad\implies\quad C=-B-2}$

$\displaystyle\longrightarrow\sf{3A+2B+C=6+2B-B-2=6+B-2=1\quad\implies\quad B=-3}$

$\displaystyle\longrightarrow\sf{C=3-2=1}$

Therefore,

$\displaystyle\longrightarrow\sf{\underline{\underline{\dfrac{x+4}{x(x+1)(x+2)}=\dfrac{2}{x}-\dfrac{3}{x+1}+\dfrac{1}{x+2}}}}$