Question

Resolve into partial fractions : $\frac{5x^3+6x^2+5x}{(x^2-1)(x+1{)}^3}$

• A. $\frac{1}{x-1}+\frac{x}{x+1}+\frac{3}{(x-1{)}^2}-\frac{3}{(x-1{)}^3}+\frac{2}{(x+1{)}^4}$
• B. $\frac{1}{x-1}-\frac{1}{x+1}+\frac{3}{(x+1{)}^2}-\frac{3}{(x+1{)}^3}+\frac{2}{(x+1{)}^4}$
• C. $\frac{6}{x-1}-\frac{3x}{(x+1{)}^2}-\frac{3}{(x+1{)}^3}-\frac{2}{(x+1{)}^4}$
• D. $\frac{1}{x-1}-\frac{1}{x+1}+\frac{3}{(x+1{)}^2}-\frac{3}{(x+1{)}^3}$

Answer 1

The correct option is B $\frac{1}{x-1}-\frac{1}{x+1}+\frac{3}{(x+1{)}^2}-\frac{3}{(x+1{)}^3}+\frac{2}{(x+1{)}^4}$

Let, $y=\frac{5x^3+6x^2+5x}{(x^2-1){(x+1)}^3}$
$y=\frac{5x^3+6x^2+5x}{(x+1)(x-1){(x+1)}^3}$
$y=\frac{5x^3+6x^2+5x}{(x-1){(x+1)}^4}$
Now, $\frac{5x^3+6x^2+5x}{(x-1){(x+1)}^4}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{{(x+1)}^2}+\frac{D}{{(x+1)}^3}+\frac{E}{{(x+1)}^4}$
$5x^3+6x^2+5x=A{(x+1)}^4+B(x-1){(x+1)}^3+C(x-1){(x+1)}^2+D(x-1)(x+1)+E(x-1)⟶1$
$\Rightarrow$ Put $x=-1$
$-5+6-5=E(-1-1)\Rightarrow E=2$
$\Rightarrow$ Put $x=1$
$5+6+5=A{(x+1)}^4\Rightarrow A=1$
$\Rightarrow$ Compare co efficient of $x^4$ , we get
$A+B=0\Rightarrow B=-1$
$\Rightarrow$ Compare co eff. of $x^3$, we get
$4A+2B+C=5\Rightarrow C=3$
$\Rightarrow$ Put $x=0$
$A-B-C-D-E=0$
$D=A-B-C-E=1+1-3-2=-3=D$
$\therefore y=\frac{1}{x-1}-\frac{1}{x+1}+\frac{3}{{(x+1)}^2}-\frac{3}{{(x+1)}^3}+\frac{2}{{(x+1)}^4}$