Question

The value of \(\int_{-1/\sqrt{2}}^{1/\sqrt{2}} \left[\left(\frac{x+1}{x-1}\right)^2 + \left(\frac{x-1}{x+1}\right)^2 - 2\right]^{1/2} dx\) is :

• A. \(4 \log_e (3 + 2\sqrt{2})\)
• B. \(\log_e 4\)
• C. \(\log_e 16\)
• D. \(2 \log_e 16\)

Hint:

Always look for perfect squares under the radical sign. Be careful with the absolute value sign when dealing with definite integrals.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The integrand is in the form of \(\sqrt{a^2 + b^2 - 2ab} = |a - b|\), where \(ab = 1\).
Step 2: Detailed Explanation:
Let \(a = \frac{x+1}{x-1}\) and \(b = \frac{x-1}{x+1}\).
The integrand is \(\sqrt{(a - b)^2} = |a - b|\).
\[ a - b = \frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{(x+1)^2 - (x-1)^2}{x^2-1} = \frac{4x}{x^2-1} \]
So the integral is \(I = \int_{-1/\sqrt{2}}^{1/\sqrt{2}} \left|\frac{4x}{x^2-1}\right| dx\).
Since the function is even (\(f(-x) = f(x)\)):
\[ I = 2 \int_0^{1/\sqrt{2}} \left|\frac{4x}{x^2-1}\right| dx \]
In the range \(0<x<1/\sqrt{2}\), \(x^2 - 1\) is negative and \(4x\) is positive. So \(\left|\frac{4x}{x^2-1}\right| = \frac{4x}{1-x^2}\).
\[ I = 2 \int_0^{1/\sqrt{2}} \frac{4x}{1-x^2} dx = 8 \int_0^{1/\sqrt{2}} \frac{x}{1-x^2} dx \]
Let \(u = 1 - x^2 \implies du = -2x dx\).
When \(x=0, u=1\); when \(x=1/\sqrt{2}, u=1/2\).
\[ I = 8 \int_1^{1/2} \frac{-du/2}{u} = 4 \int_{1/2}^1 \frac{du}{u} = 4 [\ln u]_{1/2}^1 \]
\[ I = 4 (\ln 1 - \ln(1/2)) = 4 \ln 2 = \ln 2^4 = \ln 16 \]
Step 3: Final Answer:
The value of the integral is \(\log_e 16\).