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AP EAMCET · Maths · Indefinite Integration

If \(A=\int_0^{\infty} \frac{1+x^2}{1+x^4} d x, \quad B=\int_0^1 \frac{1+x^2}{1+x^4} d x\), then

  1. A \(2 \mathrm{~A}=\mathrm{B}\)
  2. B \(\mathrm{A}=\mathrm{B}\)
  3. C \(2 \mathrm{~B}=\mathrm{A}\)
  4. D \(2 \mathrm{~B}+\mathrm{A}=0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2 \mathrm{~B}=\mathrm{A}\)

Step-by-step Solution

Detailed explanation

\(A=\int_0^{\infty} \frac{1+x^2}{1+x^4} d x, B=\int_0^1 \frac{1+x^2}{1+x^4} d x\) Let \(I=\int \frac{1+x^2}{1+x^4} d x=\int \frac{1+x^2}{\left(x^2-\sqrt{2} x+1\right)\left(x^2+\sqrt{2} x+1\right)} d x\) By partial fraction…