Answer 1

`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]