Question:
If $\frac{x^{4} + 24 x^{2} + 28}{\left(x^{2} + 1\right)^{3}} = \frac{A }{\left(x^{2} + 1\right)} + \frac{B}{\left(x^{2} + 1\right)^{2}} + \frac{C}{\left(x^{2} + 1\right)^{3}} $ then $A + C = $
If $\frac{x^{4} + 24 x^{2} + 28}{\left(x^{2} + 1\right)^{3}} = \frac{A }{\left(x^{2} + 1\right)} + \frac{B}{\left(x^{2} + 1\right)^{2}} + \frac{C}{\left(x^{2} + 1\right)^{3}} $ then $A + C = $
Updated On: Oct 23, 2024
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The Correct Option is D
Solution and Explanation
If $\frac{x^{4}+24 x^{2}+28}{\left(x^{2}+1\right)^{3}}=\frac{A}{x^{2}+1}+\frac{B}{\left(x^{2}+1\right)^{2}}$
$+\frac{C}{\left(x^{2}+1\right)^{3}}$
$\Rightarrow x^{4}+24 x^{2}+28=A\left(x^{2}+1\right)^{2}+B\left(x^{2}+1\right)+C$
On comparing the coefficient of different terms
$A=1;\, 2 A+B=24$ and $A+B+C=28$
$\Rightarrow A=1,\, B=22$, so $A+C=6$
$+\frac{C}{\left(x^{2}+1\right)^{3}}$
$\Rightarrow x^{4}+24 x^{2}+28=A\left(x^{2}+1\right)^{2}+B\left(x^{2}+1\right)+C$
On comparing the coefficient of different terms
$A=1;\, 2 A+B=24$ and $A+B+C=28$
$\Rightarrow A=1,\, B=22$, so $A+C=6$
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Concepts Used:
Integration by Partial Fractions
The number of formulas used to decompose the given improper rational functions is given below. By using the given expressions, we can quickly write the integrand as a sum of proper rational functions.
For examples,
