Question

`int_-2^1 dx/(x^2 + 4x + 13)` = ______

Options

• `pi/4`

• `pi/6`

• `pi/12`

• `pi/3`

Answer 1

`int_-2^1 dx/(x^2 + 4x + 13)` = `underline(pi/12)`

Explanation:

`int_-2^1 dx/(x^2 + 4x + 13)` = `int_-2^1dx/(x^2 + 4x + 4 + 9)`

= `int_-2^1 dx/((x + 2)^2 + (3)^2) = 1/3[tan^-1((x + 2)/3)]_-2^1`

= `1/3[tan^-1(1) - tan^-1(0)]`

= `1/3(pi/4 - 0)`

= `pi/12`