AP EAMCET · Maths · Indefinite Integration
If \(\frac{6 x^3+7 x^2+6 x-3}{(x-1)(x+3)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{B}{x+3}+\frac{C x+D}{x^2+1}\)
and \(n=A+B+C+D\) and \({ }^{50} C_n={ }^{50} C_r\), then \(r\) is equal to
- A \(40\)
- B \(43\)
- C \(35\)
- D \(42\)
Answer & Solution
Correct Answer
(D) \(42\)
Step-by-step Solution
Detailed explanation
Given, \(\frac{6 x^3+7 x^2+6 x-3}{(x-1)(x+3)\left(x^2+1\right)}\) \(=\frac{A}{x-1}+\frac{B}{x+3}+\frac{(C x+D)}{x^2+1}\) \(\Rightarrow \quad 6 x^3+7 x^2+6 x-3\) \(\begin{aligned} & =A(x+3)\left(x^2+1\right)+B(x-1)\left(x^2+1\right) \\ & +C x(x-1)(x+3)+D(x-1)(x+3)\end{aligned}\)…
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