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WBJEE · Maths · Indefinite Integration

\(\int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x(x>0)\) is

  1. A \(\tan ^{-1}\left(x+\frac{1}{x}\right)+C\)
  2. B \(\tan ^{-1}\left(x-\frac{1}{x}\right)+C\)
  3. C \(\log _{e}\left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|+C\)
  4. D \(\log _{e}\left|\frac{x-\frac{1}{x}-1}{x-\frac{1}{x}+1}\right|+C\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\tan ^{-1}\left(x+\frac{1}{x}\right)+C\)

Step-by-step Solution

Detailed explanation

Let \(I=\int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x\) \(=\int \frac{1-1 / x^{2}}{x^{2}+3+1 / x^{2}} d x\) \(=\int \frac{1-1 / x^{2}}{\left(x^{2}+\frac{1}{x^{2}}\right)+3} d x\) \(=\int \frac{1-1 / x^{2}}{\left(x+\frac{1}{x}\right)^{2}-2+3} d x\)…