resolve x²-3/(x+2)(x²+1) into partial fractions
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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)
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