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JEE Main 2022 · Mathematics · Indefinite and Definite Integrals · hard

∫(x^2-1)/(x^4+x^2+1) dx equals (where C is a constant)

Single correct option

∫x2−1x4+x2+1 dx\displaystyle\int\dfrac{x^2-1}{x^4+x^2+1}\,dx equals (where CC is a constant)
  1. A. 12ln⁡x2+x+1x2−x+1+C\dfrac12\ln\dfrac{x^2+x+1}{x^2-x+1}+C
  2. B. 12ln⁡x2−x+1x2+x+1+C\dfrac12\ln\dfrac{x^2-x+1}{x^2+x+1}+C
  3. C. ln⁡x2−x+1x2+x+1+C\ln\dfrac{x^2-x+1}{x^2+x+1}+C
  4. D. 13tan⁡−1x2−13 x+C\dfrac1{\sqrt3}\tan^{-1}\dfrac{x^2-1}{\sqrt3\,x}+C
Show answer and solution

Answer: (B) 12ln⁡x2−x+1x2+x+1+C\dfrac12\ln\dfrac{x^2-x+1}{x^2+x+1}+C

Divide numerator and denominator by x2x^2: the integrand becomes 1−1/x2x2+1+1/x2\dfrac{1-1/x^2}{x^2+1+1/x^2}. Put u=x+1xu=x+\frac1x, so du=(1−1x2)dxdu=\left(1-\frac1{x^2}\right)dx and x2+1x2+1=u2−1x^2+\frac1{x^2}+1=u^2-1. Then ∫duu2−1=12ln⁡∣u−1u+1∣+C\int\frac{du}{u^2-1}=\frac12\ln\left|\frac{u-1}{u+1}\right|+C. Substituting back, u−1u+1=x2−x+1x2+x+1\frac{u-1}{u+1}=\frac{x^2-x+1}{x^2+x+1}, which is positive for all real xx. The answer is 12ln⁡x2−x+1x2+x+1+C\frac12\ln\frac{x^2-x+1}{x^2+x+1}+C; the arctan form belongs to the numerator x2+1x^2+1.
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