The expression you have provided is:

$\frac{x^2 + 1}{(2x^2 + 1)(x^2 + 4)}$

We need to decompose this into partial fractions.

Step-by-step process:

1. Set up the partial fractions: The denominator is already factored as $(2x^2 + 1)(x^2 + 4)$, so we can express the partial fraction decomposition as:

   $\frac{x^2 + 1}{(2x^2 + 1)(x^2 + 4)} = \frac{Ax + B}{2x^2 + 1} + \frac{Cx + D}{x^2 + 4}$

   Here, $Ax + B$ and $Cx + D$ are the numerators of the respective partial fractions, where $A$, $B$, $C$, and $D$ are constants to be determined.

2. Combine the right-hand side: Multiply both sides of the equation by the common denominator $(2x^2 + 1)(x^2 + 4)$ to eliminate the denominator:

   $x^2 + 1 = (Ax + B)(x^2 + 4) + (Cx + D)(2x^2 + 1)$

3. Expand both terms: Expand the right-hand side:

   $(Ax + B)(x^2 + 4) = Ax^3 + 4Ax + Bx^2 + 4B$ $(Cx + D)(2x^2 + 1) = 2Cx^3 + Cx + 2Dx^2 + D$

   So the equation becomes:

   $x^2 + 1 = (Ax^3 + 4Ax + Bx^2 + 4B) + (2Cx^3 + Cx + 2Dx^2 + D)$

4. Group like terms: Group the powers of $x$ on the right-hand side:

   $x^2 + 1 = (A + 2C)x^3 + (B + 2D)x^2 + (4A + C)x + (4B + D)$

5. Set up a system of equations: Now, compare the coefficients of $x^3$, $x^2$, $x$, and the constant terms on both sides of the equation:

   • For $x^3$: $A + 2C = 0$
   • For $x^2$: $B + 2D = 1$
   • For $x$: $4A + C = 0$
   • For the constant: $4B + D = 1$

6. Solve the system of equations: From $A + 2C = 0$, we get $A = -2C$. Substituting into $4A + C = 0$:

   $4(-2C) + C = 0 \implies -8C + C = 0 \implies -7C = 0 \implies C = 0$

   Hence, $A = 0$.

   Now, using $B + 2D = 1$ and $4B + D = 1$:

   • $B + 2D = 1$
   • $4B + D = 1$

   Solve these simultaneously. First, multiply the first equation by 2:

   $2B + 4D = 2$

   Now subtract the second equation from this:

   $(2B + 4D) - (4B + D) = 2 - 1$ $-2B + 3D = 1 \implies D = \frac{1 + 2B}{3}$

   Substitute $D = \frac{1 + 2B}{3}$ into $B + 2D = 1$:

   $B + 2\left(\frac{1 + 2B}{3}\right) = 1$ $B + \frac{2(1 + 2B)}{3} = 1$ Multiply through by 3 to clear the denominator:

   $3B + 2(1 + 2B) = 3$ $3B + 2 + 4B = 3$ $7B + 2 = 3 \implies 7B = 1 \implies B = \frac{1}{7}$

   Now substitute $B = \frac{1}{7}$ into $B + 2D = 1$:

   $\frac{1}{7} + 2D = 1 \implies 2D = 1 - \frac{1}{7} = \frac{6}{7} \implies D = \frac{3}{7}$

Conclusion:

Thus, the partial fraction decomposition is:

$\frac{x^2 + 1}{(2x^2 + 1)(x^2 + 4)} = \frac{1/7}{x^2 + 4} + \frac{3/7}{2x^2 + 1}$

Would you like further clarification on any steps, or do you want to explore related topics?

Here are 5 related questions to consider:

1. How can partial fraction decomposition be applied to integrals?
2. What are the conditions for using partial fraction decomposition in rational functions?
3. Can this method be applied if there are repeated factors in the denominator?
4. How would you apply partial fractions in solving differential equations?
5. What happens if the degree of the numerator is higher than the denominator?

Tip: Always ensure that the degree of the numerator is less than the degree of the denominator before applying partial fractions; if not, perform polynomial long division first!