Q116 / 180 · COMEDK 2024 (Morning Shift) · Mathematics

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

  • A
    $(x-2) \sqrt{x^2-4 x+2}+\dfrac{1}{2} \log \left|(x-2)+\sqrt{x^2-4 x+2}\right|+C$
  • Correct answer
    B
    $\dfrac{1}{2}(x-2) \sqrt{x^2-4 x+2}-\log \left|(x-2)+\sqrt{x^2-4 x+2}\right|+C$
  • C
    $\dfrac{1}{2}(x-2) \sqrt{x^2-4 x+2}-\sin ^{-1} \dfrac{x-2}{\sqrt{2}}+C$
  • D
    $\dfrac{1}{2}(x-2) \sqrt{x^2-4 x+2}+\log \left|(x-2)+\sqrt{x^2-4 x+2}\right|+C$

Solution

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 $I = \int \sqrt{u^2 - 2} du$.

Using the standard formula $\int \sqrt{u^2 - a^2} du = \dfrac{u}{2} \sqrt{u^2 - a^2} - \dfrac{a^2}{2} \ln |u + \sqrt{u^2 - a^2}| + C$, where $a^2 = 2$:

$I = \dfrac{u}{2} \sqrt{u^2 - 2} - \dfrac{2}{2} \ln |u + \sqrt{u^2 - 2}| + C$.

Substituting $u = x-2$ back into the expression:

$I = \dfrac{x-2}{2} \sqrt{(x-2)^2 - 2} - \ln |(x-2) + \sqrt{(x-2)^2 - 2}| + C$.

Simplifying the expression:

$I = \dfrac{1}{2}(x-2) \sqrt{x^2 - 4x + 2} - \ln |(x-2) + \sqrt{x^2 - 4x + 2}| + C$.

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

Hence, the correct answer is option: 'B'.

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