Question

Evaluate the following:

`int_0^1 (x"d"x)/sqrt(1 + x^2)`

Answer 1

Let I = `int_0^1 (x"d"x)/sqrt(1 + x^2)`

Put 1 + x² = t

⇒ 2x dx = dt

⇒ x dx = `"dt"/2`

Changing the limits, we have

When x = 0

∴ t = 1

When x = 1

∴ t = 2

∴ I = `1/2 int_1^2 "dt"/sqrt("t")`

= `1/2 * 2["t"^(1/2)]_1^2`

= `sqrt(2) - 1`

Hence, I =  `sqrt(2) - 1`.