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COMEDK · Maths · 28. Indefinite Integration

\(\int \sqrt{x^2-4 x+2} d x=\)

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

Answer & Solution

Correct Answer

(B) \(\dfrac{1}{2}(x-2) \sqrt{x^2-4 x+2}-\log \left|(x-2)+\sqrt{x^2-4 x+2}\right|+C\)

Step-by-step Solution

Detailed explanation

The integral is \(I = \int \sqrt{x^2 - 4x + 2} dx\). Complete the square inside the square root: \(x^2 - 4x + 2 = (x^2 - 4x + 4) - 2 = (x-2)^2 - 2\). Thus, \(I = \int \sqrt{(x-2)^2 - 2} dx\). Let \(u = x-2\), then \(du = dx\). The integral becomes \(I = \int \sqrt{u^2 - 2} du\).…