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Question
Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13) "d"x`
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Solution
Let I = `int_(-4)^2 1/(x^2 + 4x + 13) "d"x`
= `int_(-4)^2 1/(x^2 + 4x + 4 + 9) "d"x`
= `int_(-4)^2 1/((x + 2)^2 + (3)^2) "d"x`
= `[1/3 tan^-1 ((x + 2)/3)]_(-4)^2`
= `1/3[tan^-1(4/3) - tan^-1(-2/3)]`
∴ I = `1/3[tan^-1(4/3) + tan-1(2/3)]` ......[∵ tan−1(− θ) = − tan−1θ]
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