The value of ∫_((- 1)/√2)^(1/√2) ( ( (x + 1)/(x - 1) )² + ( (x - 1)/(x + 1) )² - 2 )^(1/2)dx is :

Solution

Maths sir · 4 steps

The thing under the root is a perfect square. Spot it, and keep the modulus when you take the root.

1. Spot the square. Put t = (x+1)/(x-1). Then (x-1)/(x+1) = 1/t, and t² + 1/t² - 2 = (t - 1/t)². So the integrand is |t - 1/t|, not t - 1/t.

   t - 1/t = ((x+1)² - (x-1)²)/(x² - 1) = 4x/(x² - 1)

2. Remove the modulus. For |x| ≤ 1/√2 we have 1 - x² > 0, so the integrand is 4|x|/(1 - x²). This is an even function, so the integral is twice the integral from 0 to 1/√2.
3. Integrate. 2∫₀^(1/√2) 4x/(1 - x²) dx = 8[-1/2 ln(1 - x²)]₀^(1/√2) = -4 ln 1/2 = 4 ln 2.
4. So. The value is 4 logₑ 2 = logₑ 2⁴ = logₑ 16, which is option (b).

The idea to keep. When you see a² + 1/a² - 2 under a square root, it is |a - 1/a|. Always keep the modulus, then use even or odd symmetry over the limits.

Answer from the NTA final answer key. Solution by PrepWiser.