COMEDK · Maths · 28. Indefinite Integration
\(\frac{3 x^{\hat{2}}+1}{x^{2}-6 x+8}\) is equal to
- 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 \(\frac{-49}{2(x-4)}+\frac{13}{2(x-2)}\)
Answer & Solution
Correct Answer
(B) \(3+\frac{49}{2(x-4)}-\frac{13}{2(x-2)}\)
Step-by-step Solution
Detailed explanation
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…
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