Question

If $\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$, then $A+B+C+D=$

• A. 0
• B. 1
• C. -1
• D. 6

Hint:

For partial fractions involving only even powers of $x$ (like $x^2, x^4$), a quick method is to substitute $y=x^2$ and solve the simpler decomposition in terms of $y$. Then substitute back to find the coefficients.

Answer 1

The Correct Option is B

We are given the partial fraction decomposition:
$\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$.
To solve this, let's make a temporary substitution $y = x^2$. The expression becomes:
$\frac{y+1}{(y+2)(y+3)}$.
We can now decompose this simpler rational expression.
$\frac{y+1}{(y+2)(y+3)} = \frac{P}{y+2} + \frac{Q}{y+3}$.
Using the cover-up method:
To find P, cover the $(y+2)$ term and substitute $y=-2$ into the rest:
$P = \frac{-2+1}{-2+3} = \frac{-1}{1} = -1$.
To find Q, cover the $(y+3)$ term and substitute $y=-3$ into the rest:
$Q = \frac{-3+1}{-3+2} = \frac{-2}{-1} = 2$.
So, the decomposition is $\frac{-1}{y+2} + \frac{2}{y+3}$.
Now, substitute back $y = x^2$:
$\frac{-1}{x^2+2} + \frac{2}{x^2+3}$.
Comparing this with the given form $\frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$:
For the first term, $Ax+B = -1$, which implies $A=0$ and $B=-1$.
For the second term, $Cx+D = 2$, which implies $C=0$ and $D=2$.
The required value is $A+B+C+D$.
$A+B+C+D = 0 + (-1) + 0 + 2 = 1$.