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Question
Evaluate the following definite integrals:
`int_(-1)^1 ("d"x)/(x^2 + 2x + 5)`
Solution
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`
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