$\begin{aligned} & ext { If } \frac{3 x^2+1}{\left(x^2+1 ight)\left(x^2+2 ight)^2}=\frac{A x+B}{x^2+1}+\frac{C x+D}{x^2+2} \ & +\frac{E x+F}{\left(x^2+2 ight)^2}, ext { then } A+C+E=\end{aligned}$

Question

$\begin{aligned} & 	ext { If } \frac{3 x^2+1}{\left(x^2+1ight)\left(x^2+2ight)^2}=\frac{A x+B}{x^2+1}+\frac{C x+D}{x^2+2} \ & +\frac{E x+F}{\left(x^2+2ight)^2}, 	ext { then } A+C+E=\end{aligned}$

Aicharya · Accepted Answer

If $ \begin{array}{r} \frac{3 x^2+1}{\left(x^2+1ight)\left(x^2+2ight)^2}=\frac{A x+B}{x^2+1}+\frac{C x+D}{x^2+2}+\frac{E x+F}{\left(x^2+2ight)^2} \ \Rightarrow \frac{3 x^2+1}{\left(x^2+1ight)\left(x^2+2ight)^2} \ \end{array} $ $\begin{aligned} & A x+B\left(x^2+2ight)^2+(C x+D)\left(x^2+1ight) \ = & \frac{\left(x^2+2ight)+(E x+F)\left(x^2+1ight)}{\left(x^2+1ight)\left(x^2+2ight)^2}\end{aligned}$ $ \begin{gathered} \Rightarrow \quad 3 x^2+1=(A x+B)\left(x^2+2ight)^2+(C x+D) \ \left(x^2+1ight)\left(x^2+2ight)+(E x+F)\left(x^2+1ight) \end{gathered} $ On comparing the coefficient of different term, $A=0, C=0$ and $E=0$ So, $ A+C+E=0 $

$\begin{aligned} & \text { If } \frac{3 x^2+1}{\left(x^2+1\right)\left(x^2+2\right)^2}=\frac{A…

Solution