Question

The value of `int_((-1)/sqrt(2))^(1/sqrt(2)) (((x + 1)/(x - 1))^2 + ((x - 1)/(x + 1))^2 - 2)^(1/2)`dx is ______.

पर्याय

• log_e 4

• log_e 16

• `4log_e(3 + 2sqrt(2))`

• 2 log_e 16

Answer 1

The value of `int_((-1)/sqrt(2))^(1/sqrt(2)) (((x + 1)/(x - 1))^2 + ((x - 1)/(x + 1))^2 - 2)^(1/2)`dx is `underlinebb(log_e16)`.

Explanation:

I = `int_((-1)/sqrt(2))^(1/sqrt(2)) (((x + 1)/(x - 1))^2 + ((x - 1)/(x + 1))^2 - 2)^(1/2)`dx

⇒ `int_(1/sqrt(2))^((-1)/sqrt(2)) sqrt([(x - 1)/(x + 1) - (x + 1)/(x - 1)]^2) "d"x = int_((-1)/sqrt(2))^(1/sqrt(2)) sqrt(((-4x)/(x^2 - 1))^2) "d"x`

⇒ `int_((-1)/sqrt(2))^(1/sqrt(2)) sqrt((16x^2)/(1 - x^2)^2) "d"x`

⇒ `int_((-1)/sqrt(2))^(1/sqrt(2)) (4|x|)/((1 - x^2)) "d"x`,

Since f(x) = `|x|/(1 - x^2)` is an even function, 

So apply `int_(-"a")^"a" "f"|x|"d"x = 2int_0^"a""f"(x)"d"x`, we get

I = `8int_0^(1/sqrt(2)) (x"d"x)/(1 - x^2) = [-4"In"(1 - x^2)]_0^(1/sqrt(2))`

= log_e 16