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Question
Evaluate: `int_(-4)^2 (1)/(x^2 + 4x + 13)*dx`
Sum
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Solution
`int_(-4)^2 (1)/(x^2 + 4x + 13)*dx`
= `int_(-4)^2 (1)/(x^2 + 4x + 4 + 9)*dx`
= `int_(-4)^2 (1)/((x + 2)^2 + 3^2)*dx`
= `[1/3tan^-1 ((x + 2)/3)]_(-4)^2`
= `(1)/(3)tan^-1 ((2 + 2)/3) - (1)/(3)tan^-1((-4 + 2)/3)`
= `(1)/(3)tan^-1 (4/3) - (1)/(3)tan^-1 (-2/3)`
= `(1)/(3)[tan^-1 4/3 + tan^-1 2/3]` ...[∵ tan–1 (–x) = –tan–1 x]
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