Constant underestimation

$\dfrac{x^{2}+5}{(x^{2}+2)^{2}}$ = $\dfrac{1}{(x^{2}+2)}$ + $\dfrac{m}{(x^{2}+2)^{2}}$

$\dfrac{x^{2}+5}{(x^{2}+2)^{2}}$ = $\dfrac{(x^{2}+2)}{(x^{2}+2)^{2}}$ + $\dfrac{m}{(x^{2}+2)^{2}}$

$\dfrac{x^{2}+5}{(x^{2}+2)^{2}}$ = $\dfrac{(x^{2}+2)+m}{(x^{2}+2)^{2}}$

$x^{2}+5=x^{2}+2+m$

$\boxed {m=3}$

x²+5/(x²+2)²=1/(x²+2) + m/(x²+2)² x²+5/(x²+2)²=x²+2+m/(x²+2)² x²+5=x²+2+m 5-2=m 3=m m=3

$\begin{aligned} \frac {x^2+5}{(x^2+2)^2} & = \frac 1{x^2+2} + \frac m{(x^2+2)^2} & \small \color{#3D99F6} \text{Multiply both sides by }(x^2+2)^2 \\ x^2+5 & = x^2+2 + m \\ \implies m & = \boxed 3 \end{aligned}$

This question can be done by using a method used for solving integrals by using partial fractions.

we have , $\frac{x^{2}+5}{(x^{2}+2)^{2}} = \frac{1}{(x^{2}+2)} + \frac{m}{(x^{2}+2)^{2}}$

We can multiply both sides by $(x^{2}+2)^{2}$ ,doing which we get , $x^{2}+5 = x^{2}+2 + m$

solving which we get , $\boxed{m = 3}$

Take LCM on RHS, then lower part of both side will get canceled. The equation will come:x^2+5=x^2+2+m Now, x^2 will again get canceled. So, M=5 -2
=3