Question

Evaluate the following definite integrals:

`int_(-1)^1 ("d"x)/(x^2 + 2x + 5)`

Answer 1

Let I = `int_(-1)^1 ("d"x)/(x^2 + 2x + 5)`

= `int_(-1)^1 ("d"x)/((x + 1)^2 - 1 + 5)`

= `int_(-1)^1 ("d"x)/((x + 1)^2 + (2)^2)`  ........`[int ("d"x)/(x^2 + "a"^2) = 1/"a" tan^-1 (x/"a")]`

= `[1/2 tan^-1 ((x+ 1)/2)]_(-1)^1`

= `1/2[tan^-1 (2/2) - tan^-1 (0/2)]`

= `1/2[tan^-1 (1) - tan^-1 (0)]`

= `1/2 [pi/4 - 0]`

= `pi/8`