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Question: Answered & Verified by Expert
$\frac{3 x^{\hat{2}}+1}{x^{2}-6 x+8}$ is equal to
MathematicsIndefinite IntegrationCOMEDKCOMEDK 2020
Options:
  • A $\frac{49}{2(x-4)}-\frac{13}{2(x-2)} \quad$
  • B $3+\frac{49}{2(x-4)}-\frac{13}{2(x-2)}$
  • C $\frac{49}{2(x-4)}+\frac{13}{2(x-2)} \quad$
  • D $\frac{-49}{2(x-4)}+\frac{13}{2(x-2)}$
Solution:
2500 Upvotes Verified Answer
The correct answer is: $3+\frac{49}{2(x-4)}-\frac{13}{2(x-2)}$
We have,
$\frac{3 x^{2}+1}{x^{2}-6 x+8}=3+\frac{18 x-23}{x^{2}-6 x+8}$ $=3+\frac{18 x-23}{(x-4)(x-2)}$ Aguin, let $\frac{18 x-23}{(x-4)(x-2)}=\frac{A}{x-4}+\frac{B}{x-2}$ $\Rightarrow \quad 18 x-23=A(x-2)+B(x-4)$
Put $x=4$, we get
$$
72-23=2 A \Rightarrow A=\frac{49}{2}
$$
Put $x=2$, we get
$$
\begin{gathered}
36-23=-2 B \\
\Rightarrow \quad B=\frac{-13}{2} \\
\text { So, } \frac{3 x^{2}+1}{x^{2}-6 x+8}=3+\frac{49}{2(x-4)}-\frac{13}{2(x-2)}
\end{gathered}
$$

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