Solution 3:
x(x+1)25x2+20x+6=xA+x+1B+(x+1)2C
⇒5x2+20x+6=A(x+1)2+Bx(x+1)+Cx
Equating constants
6=A
Equating coefficients of x2
5=A+BB=−1
Equating coefficients of x
20=2A+B+C⇒20=12−1+C⇒C=9
∴x(x+1)25x2+20x+6=x6−x+11+(x+1)29
Solution 4:
(x+1)(x2+2x+4)x2+2x+3=(x+1)(x2+2x+4)x2+2x+4−1=x+11−(x+1)(x2+2x+4)1
Now decomposing (x+1)(x2+2x+4)1 as: x+1A+x2+2x+4Bx+C
∴(x+1)(x2+2x+4)1=x+1A+x2+2x+4Bx+C⇒1=A(x2+2x+4)+(Bx+C)(x+1)⇒1=(A+B)x2+(2A+B+C)x+4A+C
∴A+B=0;2A+B+C=0;4A+C=1⇒A=−B∴−2B+B+C=0⇒B=C
So, 4(−B)+B=1
⇒B=−31
∴A=31,C=−31
So, (x+1)(x2+2x+4)1=3(x+1)1−3(x2+2x+4)x+1
Hence, (x+1)(x2+2x+4)x2+2x+3=3(x+1)4−3(x2+2x+4)x+1
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