Question

Find the partial fraction of 3x+2/(x+1) (2x^2+3)
​

Answer 1

How do I solve the partial fraction (x^2+3x)/(x^2+1) (x+1)^2?

Note that

(x²+3x)/(x²+1)(x+1)²={(x+1)²+(x+1)-2}/(x²+1)(x+1)²=(1/x²+1)+1/(x²+1)(x+1)-2/(x²+1)(x+1)²

So we need only break down the kast two terms into partial fractions:

(i) Let 1/(x²+1)(x+1)=(ax+b)/(x²+1)+c(x+1)

1=(ax+b)(x+1)+c(x²+1)

x=-1→

c=½ ■

x=0→1=b+½→

b=½ ■

compare x²-term→0=a+c→

a=-½ ■

(ii) Let 1/(x²+1)(x+1)²=(Ax+B)/(x²+1)+C/(x+1)+D(x+1)²

1=(Ax+B)(x+1)²+C(x+1)(x²+1)+D(x²+1)

x=-1→

D=½ ■

x=0→1=B+C+½

B+C=½

x³-terms→0≈A+C→A=-C

x-term→0=A+B+C+½→0=B+½→

B=-½ ■→

C=1 ■

A=-C=-1 ■

∴Required partial fractions :

1/(x²+1)+½(1-x)/(x²+1)+½/(x+1)-2(x+½)/(x²+1)+2/(x+1)-1/(x+1)²

=½(1/(x²+1)-(5/2)(x/(x²+1)-1/(x+1)²+½/(x+1)