AP EAMCET · Maths · Indefinite Integration
If \(\frac{A}{x-a}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)\left(x^2+b^2\right)}\) then \(\mathrm{C}=\)
- A \(\frac{-1}{a^2+b^2}\)
- B \(\frac{1}{a^2+b^2}\)
- C \(\frac{-a}{a^2+b^2}\)
- D \(\frac{a}{a^2+b^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{-a}{a^2+b^2}\)
Step-by-step Solution
Detailed explanation
\(\frac{A}{(x-a)}+\frac{B x+C}{x^2+b^2}=\frac{1}{(x-a)\left(x^2+b^2\right)}\) \(\Rightarrow A\left(x^2+b^2\right)+(B x+C)(x-a)=1\) \(\Rightarrow(A+B) x^2+(C-a B) x+A b^2-a C=1\) On comparing we get, \(A+B=0\) ....(i) \(C-a B=0\) ....(ii) \(A b^2-a C=1\) ....(iii) From (i) and…
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