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26/12/2021 12:53 pm
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If integral \(\int_\frac{-1}{\sqrt{2}}^\frac{1}{\sqrt{2}}\) \(\Bigg(\Big( \frac{x-1}{x+1}\Big)^2 + \Big(\frac{x+1}{x-1}\Big)^2 -2 \Bigg)^{\frac{1}{2}}\)dx
(1) ln 16
(2) 2ln 17
(3) 3ln 16
(4) 4ln 18
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1 Answer
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26/12/2021 12:59 pm
Correct answer: (1) ln 16
Explanation:
\(\int_\frac{-1}{\sqrt{2}}^\frac{1}{\sqrt{2}}\) \(\Bigg(\Big(\frac{x-1}{x+1}\Big)^2 + \Big(\frac{x+1}{x-1}\Big)^2 -2 \Bigg)^{\frac{1}{2}}\)
= \(\int_\frac{-1}{\sqrt{2}}^\frac{1}{\sqrt{2}}\) \(|\Big(\frac{x-1}{x+1}\Big)^2 + (\Big(\frac{x+1}{x-1}\Big)|\)dx
= \(\int_\frac{-1}{\sqrt{2}}^\frac{1}{\sqrt{2}}\) \(|\frac{-4x}{x^2 - 1}|\)dx
I = 8 \(\int_0^\frac{1}{\sqrt{2}}\) \(|\frac{xdx}{1 - x^2}|\)dx
= -4\(\int_1^\frac{1}{2}\) \(|\frac{dt}{t}|\)
= 4 ln 2 = ln 16