Question

Decompose into partial fractions:$\frac{(7x+10)}{(x+1)(x^2-4)}$

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

Answer 1

The correct option is A $\frac{2}{(x-2)}-\frac{1}{(x+2)}-\frac{1}{(x+1)}$

Let $\frac{7x+10}{(x+1)(x^2-4)}=\frac{A}{(x+1)}+\frac{B}{(x-2)}+\frac{C}{(x+2)}$ ....(1)

Multiply $(x+1)(x^2-4)$ on both the sides, we get

$7x+10=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)$

Put $x=2$ to find value of B

$\therefore 12B=24\Rightarrow B=2$

Also put $x=-1$ to find value of A

$\therefore -7+10=A(-3)\left(1\right)\Rightarrow A=-1$

Now put $x=-2$ to find value of C

$\therefore -14+10=(-1)(-4)C\Rightarrow C=-1$

Substitute the values of A,B and C in equation (1), we get

$\frac{7x+10}{(x+1)(x^2-4)}=\frac{-1}{(x+1)}+\frac{2}{(x-2)}-\frac{1}{(x+2)}$