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Question 83 of 96

Q.Evaluate : ∫b∞1a2+x2 dx\displaystyle\int_{b}^{\infty}\dfrac{1}{a^2+x^2}\,dx, a>0a>0, b∈Rb\in\mathbb{R}.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2023Subjective· 2mImportance★★★★★
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Uses the standard antiderivative ∫dxa2+x2=1atan⁡−1(x/a)\int\frac{dx}{a^2+x^2}=\frac1a\tan^{-1}(x/a) and evaluates the resulting improper integral as a limit.

  1. Standard result: ∫dxa2+x2=1atan⁡−1xa+C\displaystyle\int\dfrac{dx}{a^2+x^2}=\dfrac1a\tan^{-1}\dfrac xa+C.
  2. So ∫b∞dxa2+x2=lim⁡R→∞[1atan⁡−1xa]bR=1a(lim⁡R→∞tan⁡−1Ra−tan⁡−1ba)\displaystyle\int_b^\infty\dfrac{dx}{a^2+x^2}=\lim_{R\to\infty}\left[\dfrac1a\tan^{-1}\dfrac xa\right]_b^R=\dfrac1a\left(\lim_{R\to\infty}\tan^{-1}\dfrac Ra-\tan^{-1}\dfrac ba\right). …

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