Question

solve this problem in detail​

Answer 1

Answer:

Hope this helps.

$\displaystyle\frac{1}{(x^2+9)(x^2+16)}=\frac{A}{x^2+9}+\frac{B}{x^2+16}\\ \\\Rightarrow 1 = A(x^2+16) + B(x^2+9)\\ \\\Rightarrow 1 = (A+B)x^2 + (16A+9B)$

Equating coefficients:

A + B = 0  =>  B = -A

and

16A + 9B = 1  =>  16A - 9A = 1  =>  7A = 1  =>  A = 1/7  =>  B = -1/7

So

$\displaystyle\frac{1}{(x^2+9)(x^2+16)}=\frac{1/7}{x^2+9}-\frac{1/7}{x^2+16}$