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COMEDK · Maths · 28. Indefinite Integration

\(\text { The value of } \int \dfrac{1}{x+\sqrt{x-1}} d x \text { is }\)

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

Answer & Solution

Correct Answer

(D) \(\log (x+\sqrt{x-1})-\dfrac{2}{\sqrt{3}} \tan ^{-1}\left(\dfrac{2 \sqrt{x-1}+1}{\sqrt{3}}\right)+C\)

Step-by-step Solution

Detailed explanation

Let \(I = \int \dfrac{1}{x + \sqrt{x-1}} dx\). Substitute \(t^2 = x - 1\), so \(x = t^2 + 1\) and \(dx = 2t dt\). The integral becomes \(I = \int \dfrac{2t}{t^2 + 1 + t} dt = \int \dfrac{2t}{t^2 + t + 1} dt\). Rewrite the numerator as \(2t + 1 - 1\):…