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Question: \[\int_{}^{}{\frac{dx}{\sqrt{x^{2} - 4x + 2}} =}\]...

∫dxx2−4x+2=\int_{}^{}{\frac{dx}{\sqrt{x^{2} - 4x + 2}} =}

A

log⁡∣x−2+x2+2−4x∣+c\log|x - 2 + \sqrt{x^{2} + 2 - 4x}| + c

B

log⁡∣x−2−x2+2−4x∣+c\log|x - 2 - \sqrt{x^{2} + 2 - 4x}| + c

C

log⁡∣x−2+x2+2+4x∣+c\log|x - 2 + \sqrt{x^{2} + 2 + 4x}| + c

D

log⁡∣x−2−x2+2+4x∣+c\log|x - 2 - \sqrt{x^{2} + 2 + 4x}| + c

Answer

log⁡∣x−2+x2+2−4x∣+c\log|x - 2 + \sqrt{x^{2} + 2 - 4x}| + c

Explanation

Solution

I=∫dx(x−2)2−2I = \int_{}^{}\frac{dx}{\sqrt{(x - 2)^{2} - 2}} ⇒I=∫dtt2−(2)2\Rightarrow I = \int_{}^{}\frac{dt}{\sqrt{t^{2} - (\sqrt{2})^{2}}}

Put x−2=tx - 2 = t ⇒dx=dt\Rightarrow dx = dt ⇒I=log⁡∣t+t2−2∣+c\Rightarrow I = \log|t + \sqrt{t^{2} - 2}| + c

⇒I=log⁡∣x−2+x2−4x+2∣+c\Rightarrow I = \log|x - 2 + \sqrt{x^{2} - 4x + 2}| + c