Question

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

• A. $\frac{6}{(x+2)}+\frac{6}{(x+1)}-\frac{5}{(x+1{)}^2}+\frac{3}{(x+1{)}^3}$
• B. $-\frac{6}{(x+2)}+\frac{6}{(x+1)}-\frac{5}{(x+1{)}^2}+\frac{3}{(x+1{)}^3}$
• C. $-\frac{6}{(x+2)}+\frac{3}{(x+1)}-\frac{5}{(x+1{)}^2}+\frac{3}{(x+1{)}^3}$
• D. $-\frac{3}{(x+2)}-\frac{3}{(x+1)}-\frac{5}{(x+1{)}^2}+\frac{3}{(x+1{)}^3}$

Answer 1

The correct option is B $-\frac{6}{(x+2)}+\frac{6}{(x+1)}-\frac{5}{(x+1{)}^2}+\frac{3}{(x+1{)}^3}$

Let $\frac{x^2+2}{(x+1{)}^3(x-2)}=\frac{A}{(x+1)}+\frac{B}{{(x+1)}^2}+\frac{C}{{(x+1)}^3}+\frac{D}{(x-2)}$
$\Rightarrow (x^2+2)=A{(x+1)}^2(x-2)+B(x+1)(x-2)+C(x-2)+D{(x+1)}^3$
$\Rightarrow x^2+2=A(x^3-x-2)+B(x^2-x-2)+C(x-2)+D(x^3+3x^2+3x+1)$
On comparing we get
$A+D=0,B+D=1,-A-B+C+3=0,-2A-2B-2C+D=2\Rightarrow A=6,B=-5,C=3,D=-6$
Hence
$\frac{x^2+2}{(x+1{)}^3(x-2)}=\frac{6}{(x+1)}-\frac{5}{{(x+1)}^2}+\frac{3}{{(x+1)}^3}-\frac{6}{(x-2)}$
Hence, option 'B' is correct.